Sequences & series

1. Definitions & types

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 12;17;22;g;32;12;17;22;g;32;

    Tn=5n+7Tn=5n+7

    Answer:

    g=g=

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'nn' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=5n+7
    Tn=5n+7

    For the missing term gg, we have n=4n=4:

    T4=5(4)+7g=27
    T4g=5(4)+7=27

    Submit your answer as:
  2. 6;q;16;x;26;6;q;16;x;26;

    Tn=5n+1Tn=5n+1

    Answer:
    1. q=q=
    2. x=x=
    numeric
    numeric
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=5n+1
    Tn=5n+1

    For the missing term qq, we have n=2n=2:

    T2=5(2)+1q=11
    T2q=5(2)+1=11

    For the missing term xx, we have n=4n=4:

    T4=5(4)+1x=21
    T4x=5(4)+1=21

    Submit your answer as: and

Identifying an arithmetic sequence

Given the list of numbers: 6;4;14;24;34;6;4;14;24;34;

Calculate the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d)(d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3
d=T2T1ord=T3T2ord=T4T3

Important: dd is NOT equal to T1T2T1T2 or T2T3T2T3.

Term T1T1 T2T2 T3T3 T4T4 T5T5
Value of term 6 -4 -14 -24 -34


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d)(d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=4(6)=10d=T3T2=14(4)=10d=T4T3=24(14)=10
ddd=T2T1=4(6)=10=T3T2=14(4)=10=T4T3=24(14)=10

All the values of dd are equal, which means we have found the common difference and therefore the sequence is arithmetic.


Submit your answer as:

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 39;57;75;39;57;75;

    Answer:39;57;7539;57;75; ; ;
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2
    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=57(39)=18 or d=T3T2=75(57)=18Therefore, T4=93T5=111T6=129
    d or dTherefore, T4T5T6=T2T1=57(39)=18=T3T2=75(57)=18=93=111=129

    Submit your answer as: andand
  2. 1,4;11,6;24,6;1,4;11,6;24,6;

    Answer:1,4;11,6;24,61,4;11,6;24,6; ; ;
    numeric
    numeric
    numeric
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=11,6(1,4)=13 or d=T3T2=24,6(11,6)=13Therefore, T4=37,6T5=50,6T6=63,6
    d or dTherefore, T4T5T6=T2T1=11,6(1,4)=13=T3T2=24,6(11,6)=13=37,6=50,6=63,6

    Submit your answer as: andand
  3. 12c;9c;6c;12c;9c;6c;

    Answer:12c;9c;6c12c;9c;6c; ; ;
    polynomial
    polynomial
    polynomial
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of cc:

    d=T2T1=9c(12c)=3c or d=T3T2=6c(9c)=3cTherefore, T4=3cT5=0T6=3c
    d or dTherefore, T4T5T6=T2T1=9c(12c)=3c=T3T2=6c(9c)=3c=3c=0=3c

    Submit your answer as: andand

Arithmetic sequences with algebraic terms

Consider the following arithmetic sequence:

2u4v;u4v;4v;u4v;
2u4v;u4v;4v;u4v;
  1. Find dd, the common difference of the sequence.

    Answer: d=d=
    expression
    STEP: Find dd, the common difference of the sequence.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 44 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, dd.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (dd), we subtract any two successive terms in the sequence:

    d=T2T1=(u4v)(2u4v)=u
    d=T2T1=(u4v)(2u4v)=u

    We could also have chosen the following terms:

    d=T3T2=(4v)(u4v)=u
    d=T3T2=(4v)(u4v)=u

    As expected, both calculations give the same value for dd.

    Therefore, the common difference between each term in the sequence is uu.


    Submit your answer as:
  2. Write down the next 22 terms in the sequence.

    Answer:

    T5=T5=

    T6=T6=

    expression
    expression
    STEP: Write down the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 44 terms, and the question asks us to find the next 22 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+dTn=Tn1+d. Thus to find the fifth term, add the common difference to the fourth term.

    T5=T4+u=(u4v)+u=2u4v
    T5=T4+u=(u4v)+u=2u4v

    Repeat the calculation to continue to the next term:

    T6=T5+u=(2u4v)+u=3u4v
    T6=T5+u=(2u4v)+u=3u4v

    Therefore, the required terms are: T5=2u4vT5=2u4v and T6=3u4vT6=3u4v.


    Submit your answer as: and

Arithmetic sequences: finding the common difference

  1. Calculate the common difference (dd).
    5;6;7;8;9
    5;6;7;8;9
    Answer: d=d=
    numeric
    STEP: Calculate the common difference (dd).
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the third term (T3)(T3) and the fourth term (T4)(T4). (We could also have chosen the second term (T2)(T2) and the third term (T3)(T3), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T4T3=87=1
    d=T4T3=87=1

    We can check that dd is constant by subtracting two other terms in the sequence:

    d=T3T2=76=1
    d=T3T2=76=1

    Therefore, the common difference is 11.


    Submit your answer as:
  2. Determine the next three terms of the sequence.
    Answer:

    T6=T6=
    T7=T7=
    T8=T8=

    numeric
    numeric
    numeric
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 2 / 3 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T6=T5+d=9+1=10
    T6=T5+d=9+1=10


    STEP: Calculate the following two terms
    [−2 points ⇒ 0 / 3 points left]

    Using the same method we can calculate the following two terms in the sequence:

    T7=T6+d=10+1=11T8=T7+d=11+1=12
    T7T8=T6+d=10+1=11=T7+d=11+1=12

    Therefore the sequence is,

    5;6;7;8;9;10;11;12
    5;6;7;8;9;10;11;12

    Submit your answer as: andand

Working with the common difference

  1. Find xx for the following arithmetic sequence:

    3x5+1;13x51;28x5+2;
    3x5+1;13x51;28x5+2;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: x=x=
    fraction
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 3 / 4 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2T2T1=T3T2 and we can solve for xx:

    T2T1=T3T2(13x51)(3x5+1)=(28x5+2)(13x51)
    T2T1(13x51)(3x5+1)=T3T2=(28x5+2)(13x51)

    STEP: Simplify the equation
    [−2 points ⇒ 1 / 4 points left]

    For an equation with fractions, we can multiply both sides of the equation by the LCD of the fractions to remove the denominators. In this case, the LCD is 55.

    5(13x51)5(3x5+1)=5(28x5+2)5(13x51)13x5(3x+5)=28x+10(13x5)10x10=15x+15
    5(13x51)5(3x5+1)13x5(3x+5)10x10=5(28x5+2)5(13x51)=28x+10(13x5)=15x+15

    STEP: Solve for xx
    [−1 point ⇒ 0 / 4 points left]

    Solve for xx:

    5x=25 x=5
    5x x=25=5

    Therefore, x=5x=5.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of xx. We have calculated the value of xx, so we can substitute and simplify to find the terms.


    STEP: Substitute x=5x=5 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute x=5x=5 into the expressions for each of the three terms:

    First term: T1=3x5+1=3(5)5+1=2Second term: T2=13x51=13(5)51=14Third term: T3=28x5+2=28(5)5+2=26
    First term: T1Second term: T2Third term: T3=3x5+1=3(5)5+1=2=13x51=13(5)51=14=28x5+2=28(5)5+2=26

    Therefore, the first three terms of the sequence are: 2,142,14 and 2626.


    Submit your answer as: andand

Arithmetic sequences with fractions

Given the following arithmetic sequence:

7;112;4;
7;112;4;
  1. Find the common difference, dd.

    Answer: d=d=
    numeric
    STEP: Find the common difference, dd.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 3 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, dd.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(112)(7)=32
    d=T2T1=(112)(7)=32

    Or we could also have chosen:

    d=T3T2=(4)(112)=32
    d=T3T2=(4)(112)=32

    Notice that both calculations give the same value for dd.

    Therefore the common difference is d=32d=32.


    Submit your answer as:
  2. Find the next 22 terms in the sequence.

    Answer:

    T4=T4=

    T5=T5=

    numeric
    numeric
    STEP: Find the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 22 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+dTn=Tn1+d.

    To calculate the fourth term, add the common difference to the third term:

    T4=T3+(32)=(4)32=52
    T4=T3+(32)=(4)32=52

    Continue the sequence by adding d=32d=32 again:

    T5=T4+(32)=(52)32=1
    T5=T4+(32)=(52)32=1

    Therefore, the next two terms in the sequence are: 5252 and 11.

    The full sequence is 7;112;4;52;1;12;7;112;4;52;1;12;.


    Submit your answer as: and

Common difference for an algebraic sequence

  1. Consider the following sequence:

    4y+6;3y+8;2y+10;y+12;14;
    4y+6;3y+8;2y+10;y+12;14;

    Determine the common difference (if there is one) for the terms of the sequence or write 'no common difference' in the answer box.

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tnd=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference dd at least two different ways to make sure that the values are the same:

    d=T2T1=(3y+8)(4y+6)=y+2d=T3T2=(2y+10)(3y+8)=y+2d=T4T3=(y+12)(2y+10)=y+2
    ddd=T2T1=(3y+8)(4y+6)=y+2=T3T2=(2y+10)(3y+8)=y+2=T4T3=(y+12)(2y+10)=y+2

    Each calculation gives the same answer, which means we have found the common difference: d=y+2d=y+2.


    Submit your answer as:
  2. If it is given that y=1y=1, determine the values of T1T1 and T2T2.

    Answer: T1T1 = and T2T2 =
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=4y+6T1=4y+6 and T2=3y+8T2=3y+8.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute y=1y=1 into the expressions for T1T1 and T2T2:

    T1=4y+6=4(1)+6=2T2=3y+8=3(1)+8=5
    T1T2=4y+6=4(1)+6=2=3y+8=3(1)+8=5

    Therefore, T1T1 = 2 and T2T2 = 5.


    Submit your answer as: and

Finding the common difference

Calculate the common difference (if there is one) for the list:

23;23;2;103;
23;23;2;103;

Give your answer as a fraction or write 'no common difference' in the answer box.

Answer: The common difference is: .
fraction
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1
d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate dd. Remember: this may not be an arithmetic sequence, so we need to calculate dd using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=23(23)=43d=T3T2=2(23)=43
dd=T2T1=23(23)=43=T3T2=2(23)=43

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=43d=43.


Submit your answer as:

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 9;11;13;15;17;9;11;13;15;17;

    Answer:General formula: Tn=Tn=
    T10=T10=
    T15=T15=
    T30=T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=11(9)=2Or d=T3T2=13(11)=2
    dOr d=T2T1=11(9)=2=T3T2=13(11)=2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=9Tn=a+(n1)d=9+(n1)(2)Tn=2n7
    T1TnTn=a=9=a+(n1)d=9+(n1)(2)=2n7

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=2(10)7=27T15=2(15)7=37T30=2(30)7=67
    T10T15T30=2(10)7=27=2(15)7=37=2(30)7=67

    Therefore, T10=27T10=27, T15=37T15=37 and T30=67T30=67.


    Submit your answer as: andandand
  2. 5;4;13;22;31;5;4;13;22;31;

    Answer:General formula: Tn=Tn=
    T10=T10=
    T15=T15=
    T30=T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=4(5)=9
    d=T2T1=4(5)=9

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=5Tn=a+(n1)d=5+(n1)(9)Tn=9n14
    T1TnTn=a=5=a+(n1)d=5+(n1)(9)=9n14

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=9(10)14=76T15=9(15)14=121T30=9(30)14=256
    T10T15T30=9(10)14=76=9(15)14=121=9(30)14=256

    Therefore, T10=76T10=76, T15=121T15=121 and T30=256T30=256.


    Submit your answer as: andandand

Exercises

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 6;13;20;g;34;6;13;20;g;34;

    Tn=7n+1Tn=7n+1

    Answer:

    g=g=

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'nn' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=7n+1
    Tn=7n+1

    For the missing term gg, we have n=4n=4:

    T4=−7(4)+1g=27
    T4g=−7(4)+1=27

    Submit your answer as:
  2. 14;v;32;f;50;14;v;32;f;50;

    Tn=9n5Tn=9n5

    Answer:
    1. v=v=
    2. f=f=
    numeric
    numeric
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=9n5
    Tn=9n5

    For the missing term vv, we have n=2n=2:

    T2=9(2)5v=23
    T2v=9(2)5=23

    For the missing term ff, we have n=4n=4:

    T4=9(4)5f=41
    T4f=9(4)5=41

    Submit your answer as: and

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 17;27;37;h;57;17;27;37;h;57;

    Tn=10n+7Tn=10n+7

    Answer:

    h=h=

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'nn' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=10n+7
    Tn=10n+7

    For the missing term hh, we have n=4n=4:

    T4=10(4)+7h=47
    T4h=10(4)+7=47

    Submit your answer as:
  2. 4;q;4;w;12;4;q;4;w;12;

    Tn=4n8Tn=4n8

    Answer:
    1. q=q=
    2. w=w=
    numeric
    numeric
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=4n8
    Tn=4n8

    For the missing term qq, we have n=2n=2:

    T2=4(2)8q=0
    T2q=4(2)8=0

    For the missing term ww, we have n=4n=4:

    T4=4(4)8w=8
    T4w=4(4)8=8

    Submit your answer as: and

Completing a sequence using the general formula

The general term is given for each sequence below. Calculate the missing term(s).

  1. 0;7;14;h;28;0;7;14;h;28;

    Tn=7n+7Tn=7n+7

    Answer:

    h=h=

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What is the position of the term that you want to find? The position is the value of 'nn' that we substitute into the general formula.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    To find the missing term, we use the equation for the general term:

    Tn=7n+7
    Tn=7n+7

    For the missing term hh, we have n=4n=4:

    T4=−7(4)+7h=21
    T4h=−7(4)+7=21

    Submit your answer as:
  2. 8;t;30;z;52;8;t;30;z;52;

    Tn=11n3Tn=11n3

    Answer:
    1. t=t=
    2. z=z=
    numeric
    numeric
    STEP: <no title>
    [−4 points ⇒ 0 / 4 points left]

    To find the two missing terms, we use the equation for the general term:

    Tn=11n3
    Tn=11n3

    For the missing term tt, we have n=2n=2:

    T2=11(2)3t=19
    T2t=11(2)3=19

    For the missing term zz, we have n=4n=4:

    T4=11(4)3z=41
    T4z=11(4)3=41

    Submit your answer as: and

Identifying an arithmetic sequence

Given the list of numbers: 4;10;16;22;28;4;10;16;22;28;

Calculate the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d)(d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3
d=T2T1ord=T3T2ord=T4T3

Important: dd is NOT equal to T1T2T1T2 or T2T3T2T3.

Term T1T1 T2T2 T3T3 T4T4 T5T5
Value of term 4 10 16 22 28


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d)(d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=10(4)=6d=T3T2=16(10)=6d=T4T3=22(16)=6
ddd=T2T1=10(4)=6=T3T2=16(10)=6=T4T3=22(16)=6

All the values of dd are equal, which means we have found the common difference and therefore the sequence is arithmetic.


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Identifying an arithmetic sequence

Given the following list of numbers: 5;14;23;32;41;5;14;23;32;41;

Determine the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d)(d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3
d=T2T1ord=T3T2ord=T4T3

Important: dd is NOT equal to T1T2T1T2 or T2T3T2T3.

Term T1T1 T2T2 T3T3 T4T4 T5T5
Value of term 5 14 23 32 41


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d)(d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=14(5)=9d=T3T2=23(14)=9d=T4T3=32(23)=9
ddd=T2T1=14(5)=9=T3T2=23(14)=9=T4T3=32(23)=9

All the values of dd are equal, which means we have found the common difference and therefore the sequence is arithmetic.


Submit your answer as:

Identifying an arithmetic sequence

Given the following list of numbers: 16;3;5;23;33;16;3;5;23;33;

Find the common difference (if there is one).

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The common difference is the difference between any two successive terms of an arithmetic sequence. If a sequence does not have a common difference between successive terms, then it is not arithmetic.
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

The common difference (d)(d) is calculated by finding the difference between any two successive terms:

d=T2T1ord=T3T2ord=T4T3
d=T2T1ord=T3T2ord=T4T3

Important: dd is NOT equal to T1T2T1T2 or T2T3T2T3.

Term T1T1 T2T2 T3T3 T4T4 T5T5
Value of term 16 -3 -5 -23 -33


STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We calculate the common difference (d)(d) at least two different ways (to show that it is constant throughout the sequence):

d=T2T1=3(16)=19d=T3T2=5(3)=2d=T4T3=23(5)=18
ddd=T2T1=3(16)=19=T3T2=5(3)=2=T4T3=23(5)=18

We see that the results are not the same - the difference between terms is not 'common.' Therefore, the sequence is not arithmetic.


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Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 15;13;11;15;13;11;

    Answer:15;13;1115;13;11; ; ;
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2
    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=13(15)=2 or d=T3T2=11(13)=2Therefore, T4=9T5=7T6=5
    d or dTherefore, T4T5T6=T2T1=13(15)=2=T3T2=11(13)=2=9=7=5

    Submit your answer as: andand
  2. 17,3;7,3;2,7;17,3;7,3;2,7;

    Answer:17,3;7,3;2,717,3;7,3;2,7; ; ;
    numeric
    numeric
    numeric
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=7,3(17,3)=10 or d=T3T2=2,7(7,3)=10Therefore, T4=12,7T5=22,7T6=32,7
    d or dTherefore, T4T5T6=T2T1=7,3(17,3)=10=T3T2=2,7(7,3)=10=12,7=22,7=32,7

    Submit your answer as: andand
  3. 18t;4t;10t;18t;4t;10t;

    Answer:18t;4t;10t18t;4t;10t; ; ;
    polynomial
    polynomial
    polynomial
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of tt:

    d=T2T1=4t(18t)=14t or d=T3T2=10t(4t)=14tTherefore, T4=24tT5=38tT6=52t
    d or dTherefore, T4T5T6=T2T1=4t(18t)=14t=T3T2=10t(4t)=14t=24t=38t=52t

    Submit your answer as: andand

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 3;9;21;3;9;21;

    Answer:3;9;213;9;21; ; ;
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2
    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=9(3)=12 or d=T3T2=21(9)=12Therefore, T4=33T5=45T6=57
    d or dTherefore, T4T5T6=T2T1=9(3)=12=T3T2=21(9)=12=33=45=57

    Submit your answer as: andand
  2. 18,6;23,6;28,6;18,6;23,6;28,6;

    Answer:18,6;23,6;28,618,6;23,6;28,6; ; ;
    numeric
    numeric
    numeric
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=23,6(18,6)=5 or d=T3T2=28,6(23,6)=5Therefore, T4=33,6T5=38,6T6=43,6
    d or dTherefore, T4T5T6=T2T1=23,6(18,6)=5=T3T2=28,6(23,6)=5=33,6=38,6=43,6

    Submit your answer as: andand
  3. 18p;38p;58p;18p;38p;58p;

    Answer:18p;38p;58p18p;38p;58p; ; ;
    polynomial
    polynomial
    polynomial
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of pp:

    d=T2T1=38p(18p)=20p or d=T3T2=58p(38p)=20pTherefore, T4=78pT5=98pT6=118p
    d or dTherefore, T4T5T6=T2T1=38p(18p)=20p=T3T2=58p(38p)=20p=78p=98p=118p

    Submit your answer as: andand

Completing the sequence

Write down the next three terms in each of the following arithmetic sequences:

  1. 49;58;67;49;58;67;

    Answer:49;58;6749;58;67; ; ;
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the common difference:

    d=T2T1 or d=T3T2
    d=T2T1 or d=T3T2

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=58(49)=9 or d=T3T2=67(58)=9Therefore, T4=76T5=85T6=94
    d or dTherefore, T4T5T6=T2T1=58(49)=9=T3T2=67(58)=9=76=85=94

    Submit your answer as: andand
  2. 11,6;19,6;27,6;11,6;19,6;27,6;

    Answer:11,6;19,6;27,611,6;19,6;27,6; ; ;
    numeric
    numeric
    numeric
    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    To complete the sequence, we need to determine the common difference:

    d=T2T1=19,6(11,6)=8 or d=T3T2=27,6(19,6)=8Therefore, T4=35,6T5=43,6T6=51,6
    d or dTherefore, T4T5T6=T2T1=19,6(11,6)=8=T3T2=27,6(19,6)=8=35,6=43,6=51,6

    Submit your answer as: andand
  3. 26w;21w;16w;26w;21w;16w;

    Answer:26w;21w;16w26w;21w;16w; ; ;
    polynomial
    polynomial
    polynomial
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    This question involves variables, but the same method is used to find the common difference and to complete the sequence as the previous two examples.
    STEP: <no title>
    [−6 points ⇒ 0 / 6 points left]

    To complete the sequence, we need to determine the common difference in terms of ww:

    d=T2T1=21w(26w)=5w or d=T3T2=16w(21w)=5wTherefore, T4=11wT5=6wT6=w
    d or dTherefore, T4T5T6=T2T1=21w(26w)=5w=T3T2=16w(21w)=5w=11w=6w=w

    Submit your answer as: andand

Arithmetic sequences with algebraic terms

Consider the following arithmetic sequence:

13qs;15qs;17qs;19qs;
13qs;15qs;17qs;19qs;
  1. Calculate the common difference (dd).

    Answer: d=d=
    expression
    STEP: Calculate the common difference (dd).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 44 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, dd.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (dd), we subtract any two successive terms in the sequence:

    d=T2T1=(15qs)(13qs)=2q
    d=T2T1=(15qs)(13qs)=2q

    We could also have chosen the following terms:

    d=T3T2=(17qs)(15qs)=2q
    d=T3T2=(17qs)(15qs)=2q

    As expected, both calculations give the same value for dd.

    Therefore, the common difference between each term in the sequence is 2q2q.


    Submit your answer as:
  2. Calculate the next 22 terms in the sequence.

    Answer:

    T5=T5=

    T6=T6=

    expression
    expression
    STEP: Calculate the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 44 terms, and the question asks us to find the next 22 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+dTn=Tn1+d. Thus to find the fifth term, add the common difference to the fourth term.

    T5=T4+2q=(19qs)+2q=21qs
    T5=T4+2q=(19qs)+2q=21qs

    Repeat the calculation to continue to the next term:

    T6=T5+2q=(21qs)+2q=23qs
    T6=T5+2q=(21qs)+2q=23qs

    Therefore, the required terms are: T5=21qsT5=21qs and T6=23qsT6=23qs.


    Submit your answer as: and

Arithmetic sequences with algebraic terms

The following arithmetic sequence is given:

k+10s;k+15s;k+20s;
k+10s;k+15s;k+20s;
  1. Find dd, the common difference of the sequence.

    Answer: d=d=
    expression
    STEP: Find dd, the common difference of the sequence.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 33 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, dd.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (dd), we subtract any two successive terms in the sequence:

    d=T2T1=(k+15s)(k+10s)=5s
    d=T2T1=(k+15s)(k+10s)=5s

    We could also have chosen the following terms:

    d=T3T2=(k+20s)(k+15s)=5s
    d=T3T2=(k+20s)(k+15s)=5s

    As expected, both calculations give the same value for dd.

    Therefore, the common difference between each term in the sequence is 5s5s.


    Submit your answer as:
  2. Write down the next 22 terms in the sequence.

    Answer:

    T4=T4=

    T5=T5=

    expression
    expression
    STEP: Write down the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 33 terms, and the question asks us to find the next 22 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+dTn=Tn1+d. Thus to find the fourth term, add the common difference to the third term.

    T4=T3+5s=(k+20s)+5s=k+25s
    T4=T3+5s=(k+20s)+5s=k+25s

    Repeat the calculation to continue to the next term:

    T5=T4+5s=(k+25s)+5s=k+30s
    T5=T4+5s=(k+25s)+5s=k+30s

    Therefore, the required terms are: T4=k+25sT4=k+25s and T5=k+30sT5=k+30s.


    Submit your answer as: and

Arithmetic sequences with algebraic terms

For the following arithmetic sequence:

s4z;5s4z;9s4z;
s4z;5s4z;9s4z;
  1. Calculate the common difference (dd).

    Answer: d=d=
    expression
    STEP: Calculate the common difference (dd).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 33 terms of a sequence and specifically tells us that the terms follow an arithmetic pattern. We must find the common difference of the sequence, dd.

    This is an example of a non-numeric arithmetic sequence, where each term in the given sequence is an algebraic expression. To find the common difference (dd), we subtract any two successive terms in the sequence:

    d=T2T1=(5s4z)(s4z)=4s
    d=T2T1=(5s4z)(s4z)=4s

    We could also have chosen the following terms:

    d=T3T2=(9s4z)(5s4z)=4s
    d=T3T2=(9s4z)(5s4z)=4s

    As expected, both calculations give the same value for dd.

    Therefore, the common difference between each term in the sequence is 4s4s.


    Submit your answer as:
  2. Find the next 22 terms in the sequence.

    Answer:

    T4=T4=

    T5=T5=

    expression
    expression
    STEP: Find the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    We have the common difference of the sequence and the first 33 terms, and the question asks us to find the next 22 terms.

    To determine any term in the sequence, we add the common difference to the term before it. In other words: Tn=Tn1+dTn=Tn1+d. Thus to find the fourth term, add the common difference to the third term.

    T4=T3+4s=(9s4z)+4s=13s4z
    T4=T3+4s=(9s4z)+4s=13s4z

    Repeat the calculation to continue to the next term:

    T5=T4+4s=(13s4z)+4s=17s4z
    T5=T4+4s=(13s4z)+4s=17s4z

    Therefore, the required terms are: T4=13s4zT4=13s4z and T5=17s4zT5=17s4z.


    Submit your answer as: and

Arithmetic sequences: finding the common difference

  1. Calculate the common difference (dd).
    10;2;6
    10;2;6
    Answer: d=d=
    numeric
    STEP: Calculate the common difference (dd).
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the second term (T2)(T2) and the third term (T3)(T3). (We could also have chosen the first term (T1)(T1) and the second term (T2)(T2), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T3T2=62=8
    d=T3T2=62=8

    We can check that dd is constant by subtracting two other terms in the sequence:

    d=T2T1=210=8
    d=T2T1=210=8

    Therefore, the common difference is 88.


    Submit your answer as:
  2. Write down the next three terms of the sequence.
    Answer:

    T4=T4=
    T5=T5=
    T6=T6=

    numeric
    numeric
    numeric
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 2 / 3 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T4=T3+d=68=14
    T4=T3+d=68=14


    STEP: Calculate the following two terms
    [−2 points ⇒ 0 / 3 points left]

    Using the same method we can calculate the following two terms in the sequence:

    T5=T4+d=148=22T6=T5+d=228=30
    T5T6=T4+d=148=22=T5+d=228=30

    Therefore the sequence is,

    10;2;6;14;22;30
    10;2;6;14;22;30

    Submit your answer as: andand

Arithmetic sequences: finding the common difference

  1. Determine the common difference.
    9;13;17;21;25
    9;13;17;21;25
    Answer: d=d=
    numeric
    STEP: Determine the common difference.
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the fourth term (T4)(T4) and the fifth term (T5)(T5). (We could also have chosen the third term (T3)(T3) and the fourth term (T4)(T4), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T5T4=25(21)=25+21=4
    d=T5T4=25(21)=25+21=4

    We can check that dd is constant by subtracting two other terms in the sequence:

    d=T4T3=21(17)=21+17=4
    d=T4T3=21(17)=21+17=4

    Therefore, the common difference is 44.


    Submit your answer as:
  2. Find the next three terms of the arithmetic sequence.
    Answer:

    T6=T6=
    T7=T7=
    T8=T8=

    numeric
    numeric
    numeric
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 2 / 3 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T6=T5+d=254=29
    T6=T5+d=254=29


    STEP: Calculate the following two terms
    [−2 points ⇒ 0 / 3 points left]

    Using the same method we can calculate the following two terms in the sequence:

    T7=T6+d=294=33T8=T7+d=334=37
    T7T8=T6+d=294=33=T7+d=334=37

    Therefore the sequence is,

    9;13;17;21;25;29;33;37
    9;13;17;21;25;29;33;37

    Submit your answer as: andand

Arithmetic sequences: finding the common difference

Consider the following sequence:

20;21;22
20;21;22
  1. Find the common difference.
    Answer: d=d=
    numeric
    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    To find the common difference, we subtract any two successive terms in the sequence. Let us use the second term (T2)(T2) and the third term (T3)(T3). (We could also have chosen the first term (T1)(T1) and the second term (T2)(T2), both should give us the same answer.)

    We find the difference between any term in the sequence and the term before it:

    d=T3T2=2221=1
    d=T3T2=2221=1

    We can check that dd is constant by subtracting two other terms in the sequence:

    d=T2T1=2120=1
    d=T2T1=2120=1

    Therefore, the common difference is 11.


    Submit your answer as:
  2. Find the next four terms of the arithmetic sequence.
    Answer:

    T4=T4=
    T5=T5=
    T6=T6=
    T7=T7=

    numeric
    numeric
    numeric
    numeric
    STEP: Calculate the next term in the sequence
    [−1 point ⇒ 3 / 4 points left]

    Add the common difference to the last known term to get the next term in the sequence:

    T4=T3+d=22+1=23
    T4=T3+d=22+1=23


    STEP: Calculate the following three terms
    [−3 points ⇒ 0 / 4 points left]

    Using the same method we can calculate the following three terms in the sequence:

    T5=T4+d=23+1=24T6=T5+d=24+1=25T7=T6+d=25+1=26
    T5T6T7=T4+d=23+1=24=T5+d=24+1=25=T6+d=25+1=26

    Therefore the sequence is,

    20;21;22;23;24;25;26
    20;21;22;23;24;25;26

    Submit your answer as: andandand

Working with the common difference

  1. Calculate xx for the following arithmetic sequence:

    2x2;x;2x+2;
    2x2;x;2x+2;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: x=x=
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 2 / 3 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2T2T1=T3T2 and we can solve for xx:

    T2T1=T3T2(x)(2x2)=(2x+2)(x)
    T2T1(x)(2x2)=T3T2=(2x+2)(x)

    STEP: Simplify the equation
    [−1 point ⇒ 1 / 3 points left]

    Collect like terms to simplify each side of the equation:

    x+2=x+2
    x+2=x+2

    STEP: Solve for xx
    [−1 point ⇒ 0 / 3 points left]

    Solve for xx:

    0=2x 0=x
    0 0=2x=x

    Therefore, x=0x=0.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of xx. We have calculated the value of xx, so we can substitute and simplify to find the terms.


    STEP: Substitute x=0x=0 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute x=0x=0 into the expressions for each of the three terms:

    First term: T1=2x2=2(0)2=2Second term: T2=x=(0)=0Third term: T3=2x+2=2(0)+2=2
    First term: T1Second term: T2Third term: T3=2x2=2(0)2=2=x=(0)=0=2x+2=2(0)+2=2

    Therefore, the first three terms of the sequence are: 2,02,0 and 22.


    Submit your answer as: andand

Working with the common difference

  1. Find zz for the following arithmetic sequence:

    4z553;11z523;21z523;
    4z553;11z523;21z523;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: z=z=
    fraction
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 3 / 4 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2T2T1=T3T2 and we can solve for zz:

    T2T1=T3T2(11z523)(4z553)=(21z523)(11z523)
    T2T1(11z523)(4z553)=T3T2=(21z523)(11z523)

    STEP: Simplify the equation
    [−2 points ⇒ 1 / 4 points left]

    For an equation with fractions, we can multiply both sides of the equation by the LCD of the fractions to remove the denominators. In this case, the LCD is 1515.

    15(11z523)15(4z553)=15(21z523)15(11z523)33z10(12z25)=63z10(33z10)45z+15=30z
    15(11z523)15(4z553)33z10(12z25)45z+15=15(21z523)15(11z523)=63z10(33z10)=30z

    STEP: Solve for zz
    [−1 point ⇒ 0 / 4 points left]

    Solve for zz:

    15=15z 1=z
    15 1=15z=z

    Therefore, z=1z=1.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of zz. We have calculated the value of zz, so we can substitute and simplify to find the terms.


    STEP: Substitute z=1z=1 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute z=1z=1 into the expressions for each of the three terms:

    First term: T1=4z553=4(1)553=1315Second term: T2=11z523=11(1)523=4315Third term: T3=21z523=21(1)523=7315
    First term: T1Second term: T2Third term: T3=4z553=4(1)553=1315=11z523=11(1)523=4315=21z523=21(1)523=7315

    Therefore, the first three terms of the sequence are: 1315,43151315,4315 and 73157315.


    Submit your answer as: andand

Working with the common difference

  1. Compute the value of yy for the following arithmetic sequence:

    y1;3y1;7y3;
    y1;3y1;7y3;

    If the answer is a non-integer, write the answer as a simplified fraction.

    Answer: y=y=
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The key to starting this question is the fact that the terms form an arithmetic sequence: that means that the difference between the first and second terms is equal to the difference between the second and third terms.


    STEP: Set up an equation based on the common difference of the terms
    [−1 point ⇒ 2 / 3 points left]

    The given sequence is arithmetic, so the difference between successive terms will be the same (common). So we know that T2T1=T3T2T2T1=T3T2 and we can solve for yy:

    T2T1=T3T2(3y1)(y1)=(7y3)(3y1)
    T2T1(3y1)(y1)=T3T2=(7y3)(3y1)

    STEP: Simplify the equation
    [−1 point ⇒ 1 / 3 points left]

    Collect like terms to simplify each side of the equation:

    2y=4y2
    2y=4y2

    STEP: Solve for yy
    [−1 point ⇒ 0 / 3 points left]

    Solve for yy:

    2y=2 y=1
    2y y=2=1

    Therefore, y=1y=1.


    Submit your answer as:
  2. Determine the numeric value of the first three terms. If the answers are not integers, write your answers as fractions, not decimals.

    Answer: ; ; ;
    fraction
    fraction
    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The first three terms in the sequence are given in terms of yy. We have calculated the value of yy, so we can substitute and simplify to find the terms.


    STEP: Substitute y=1y=1 into the expressions for the terms
    [−3 points ⇒ 0 / 3 points left]

    To find the values of the terms, we need to substitute y=1y=1 into the expressions for each of the three terms:

    First term: T1=y1=(1)1=0Second term: T2=3y1=3(1)1=2Third term: T3=7y3=7(1)3=4
    First term: T1Second term: T2Third term: T3=y1=(1)1=0=3y1=3(1)1=2=7y3=7(1)3=4

    Therefore, the first three terms of the sequence are: 0,20,2 and 44.


    Submit your answer as: andand

Arithmetic sequences with fractions

Given the following arithmetic sequence:

5;113;73;1;13;
5;113;73;1;13;
  1. Determine the common difference, dd.

    Answer: d=d=
    numeric
    STEP: Determine the common difference, dd.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 5 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, dd.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(113)(5)=43
    d=T2T1=(113)(5)=43

    Or we could also have chosen:

    d=T3T2=(73)(113)=43
    d=T3T2=(73)(113)=43

    Notice that both calculations give the same value for dd.

    Therefore the common difference is d=43d=43.


    Submit your answer as:
  2. Determine the next 22 terms in the sequence.

    Answer:

    T6=T6=

    T7=T7=

    numeric
    numeric
    STEP: Determine the next 22 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 22 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+dTn=Tn1+d.

    To calculate the sixth term, add the common difference to the fifth term:

    T6=T5+(43)=(13)43=53
    T6=T5+(43)=(13)43=53

    Continue the sequence by adding d=43d=43 again:

    T7=T6+(43)=(53)43=3
    T7=T6+(43)=(53)43=3

    Therefore, the next two terms in the sequence are: 5353 and 33.

    The full sequence is 5;113;73;1;13;53;3;133;.


    Submit your answer as: and

Arithmetic sequences with fractions

Consider the following arithmetic sequence:

2;72;5;
  1. Calculate the common difference (d).

    Answer: d=
    numeric
    STEP: Calculate the common difference (d).
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 3 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, d.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(72)(2)=32

    Or we could also have chosen:

    d=T3T2=(5)(72)=32

    Notice that both calculations give the same value for d.

    Therefore the common difference is d=32.


    Submit your answer as:
  2. Determine the next 2 terms in the sequence.

    Answer:

    T4=

    T5=

    numeric
    numeric
    STEP: Determine the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 2 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+d.

    To calculate the fourth term, add the common difference to the third term:

    T4=T3+(32)=(5)32=132

    Continue the sequence by adding d=32 again:

    T5=T4+(32)=(132)32=8

    Therefore, the next two terms in the sequence are: 132 and 8.

    The full sequence is 2;72;5;132;8;192;.


    Submit your answer as: and

Arithmetic sequences with fractions

The following arithmetic sequence is given:

5;285;315;345;
  1. Determine the common difference, d.

    Answer: d=
    numeric
    STEP: Determine the common difference, d.
    [−1 point ⇒ 0 / 1 points left]

    The question gives us 4 terms of a sequence and specifically tells us that the numbers form an arithmetic pattern. We need to find the common difference of that pattern, d.

    To determine the common difference for an arithmetic sequence, we subtract any two successive terms in the sequence: d=TnTn1. In other words the common difference is the difference between any term in the sequence and the term before it:

    d=T2T1=(285)(5)=35

    Or we could also have chosen:

    d=T3T2=(315)(285)=35

    Notice that both calculations give the same value for d.

    Therefore the common difference is d=35.


    Submit your answer as:
  2. Find the next 2 terms in the sequence.

    Answer:

    T5=

    T6=

    numeric
    numeric
    STEP: Find the next 2 terms in the sequence.
    [−2 points ⇒ 0 / 2 points left]

    Now we must find the next 2 terms in the sequence. We can do this using the common difference: any term in the sequence is equal to the term before it plus the common difference: Tn=Tn1+d.

    To calculate the fifth term, add the common difference to the fourth term:

    T5=T4+35=(345)+35=375

    Continue the sequence by adding d=35 again:

    T6=T5+35=(375)+35=8

    Therefore, the next two terms in the sequence are: 375 and 8.

    The full sequence is 5;285;315;345;375;8;435;.


    Submit your answer as: and

Common difference for an algebraic sequence

  1. Consider the following sequence:

    3y4;y3;2y3;5y1;8y;

    Determine the common difference (if there is one) for the terms of the sequence or write 'no common difference' in the answer box.

    Answer:
    string
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(y3)(3y4)=2y+1d=T3T2=(2y3)(y3)=3yd=T4T3=(5y1)(2y3)=3y+2

    You can see that the results are not equal - the difference is not 'common'. Therefore, the sequence is not arithmetic.


    Submit your answer as:
  2. If it is given that y=3, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=3y4 and T2=y3.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute y=3 into the expressions for T1 and T2:

    T1=3y4=3(3)4=5T2=y3=(3)3=0

    Therefore, T1 = 5 and T2 = 0.


    Submit your answer as: and

Common difference for an algebraic sequence

  1. Consider the following list:

    2x3;4x2;6x1;8x;10x+1;

    Calculate the common difference (if there is one) for the terms of the list or write 'no common difference' in the answer box.

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(4x2)(2x3)=2x+1d=T3T2=(6x1)(4x2)=2x+1d=T4T3=(8x)(6x1)=2x+1

    Each calculation gives the same answer, which means we have found the common difference: d=2x+1.


    Submit your answer as:
  2. If it is given that x=1, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=2x3 and T2=4x2.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute x=1 into the expressions for T1 and T2:

    T1=2x3=2(1)3=5T2=4x2=4(1)2=6

    Therefore, T1 = -5 and T2 = -6.


    Submit your answer as: and

Common difference for an algebraic sequence

  1. Consider the following pattern:

    4x+4;5x+1;6x2;7x5;8x8;

    Determine the common difference (if there is one) for the terms of the pattern or write 'no common difference' in the answer box.

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The common difference for terms in an arithmetic pattern: d=Tn+1Tn.


    STEP: Find the common difference.
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the common difference d at least two different ways to make sure that the values are the same:

    d=T2T1=(5x+1)(4x+4)=x3d=T3T2=(6x2)(5x+1)=x3d=T4T3=(7x5)(6x2)=x3

    Each calculation gives the same answer, which means we have found the common difference: d=x3.


    Submit your answer as:
  2. If it is given that x=2, determine the values of T1 and T2.

    Answer: T1 = and T2 =
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    From the question we know that T1=4x+4 and T2=5x+1.
    STEP: Substitute the given value into the expression for each term.
    [−2 points ⇒ 0 / 2 points left]

    Substitute x=2 into the expressions for T1 and T2:

    T1=4x+4=4(2)+4=12T2=5x+1=5(2)+1=11

    Therefore, T1 = 12 and T2 = 11.


    Submit your answer as: and

Finding the common difference

Calculate the common difference (if there is one) for the following list of numbers:

43;3;143;193;

Give your answer as a fraction or write 'no common difference' in the answer box.

Answer: The common difference is: .
fraction
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=3(43)=53d=T3T2=143(3)=53

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=53.


Submit your answer as:

Finding the common difference

Determine the common difference (if there is one) for the list:

1,01;1,63;2,25;2,87;

Give your answer as a decimal or write 'no common difference' in the answer box.

Answer: The common difference is: .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=1,63(1,01)=0,62d=T3T2=2,25(1,63)=0,62

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=0,62.


Submit your answer as:

Finding the common difference

Determine the common difference (if there is one) for the following list of numbers:

1,77;2,1;2,43;2,76;

Give your answer as a decimal or write 'no common difference' in the answer box.

Answer: The common difference is: .
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

We can check if there is a common difference between successive terms by calculating:

d=TnTn1

STEP: Calculate the difference between consecutive terms
[−1 point ⇒ 0 / 1 points left]

Use any two consecutive terms to calculate d. Remember: this may not be an arithmetic sequence, so we need to calculate d using at least two different pairs of terms to see whether or not the values are the same:

d=T2T1=2,1(1,77)=0,33d=T3T2=2,43(2,1)=0,33

In this case, we see that the two answers are the same, which means that the sequence of numbers is arithmetic. Therefore, the common difference is d=0,33.


Submit your answer as:

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 9;16;23;30;37;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=16(9)=7Or d=T3T2=23(16)=7

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=9Tn=a+(n1)d=9+(n1)(7)Tn=7n+2

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=7(10)+2=72T15=7(15)+2=107T30=7(30)+2=212

    Therefore, T10=72, T15=107 and T30=212.


    Submit your answer as: andandand
  2. 11;5;1;7;13;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=5(11)=6

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=11Tn=a+(n1)d=11+(n1)(6)Tn=6n+17

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=6(10)+17=43T15=6(15)+17=73T30=6(30)+17=163

    Therefore, T10=43, T15=73 and T30=163.


    Submit your answer as: andandand

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 21;27;33;39;45;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=27(21)=6Or d=T3T2=33(27)=6

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=21Tn=a+(n1)d=21+(n1)(6)Tn=6n+15

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=6(10)+15=75T15=6(15)+15=105T30=6(30)+15=195

    Therefore, T10=75, T15=105 and T30=195.


    Submit your answer as: andandand
  2. 12;19;26;33;40;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=19(12)=7

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=12Tn=a+(n1)d=12+(n1)(7)Tn=7n+5

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=7(10)+5=75T15=7(15)+5=110T30=7(30)+5=215

    Therefore, T10=75, T15=110 and T30=215.


    Submit your answer as: andandand

Arithmetic sequences: finding the general formula

Determine the general formula and then calculate T10, T15 and T30 for the following sequences:

  1. 12;14;16;18;20;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    Find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Calculate the common difference:

    d=T2T1=14(12)=2Or d=T3T2=16(14)=2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Determine the general formula:

    T1=a=12Tn=a+(n1)d=12+(n1)(2)Tn=2n10

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Find the three terms:

    T10=2(10)10=30T15=2(15)10=40T30=2(30)10=70

    Therefore, T10=30, T15=40 and T30=70.


    Submit your answer as: andandand
  2. 19;26;33;40;47;

    Answer:General formula: Tn=
    T10=
    T15=
    T30=
    polynomial
    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]
    We follow the same steps as the first part of the question: find the common difference and use it to determine the general formula.
    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    Find the common difference:

    d=T2T1=26(19)=7

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now determine the general formula:

    T1=a=19Tn=a+(n1)d=19+(n1)(7)Tn=7n+12

    STEP: <no title>
    [−3 points ⇒ 0 / 6 points left]

    Finally, calculate the three required terms:

    T10=7(10)+12=82T15=7(15)+12=117T30=7(30)+12=222

    Therefore, T10=82, T15=117 and T30=222.


    Submit your answer as: andandand

2. Arithmetic sequences

Exercises

3. Arithmetic series

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=42814

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=4.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=2(4)=6

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(4)+(k1)(6)=46k+6=6k+10

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=6
    2. c=10

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=yx(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=nk=1(6k+10)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(4)+(n1)(6)]=n2[86n+6]=n2[6n+14]=3n2+7n

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=3+5+13+21+

to n terms. Sn is also described by the expression

Sn=4n27n

Another sequence is defined as:

Q1=7Q2=73Q3=73+5Q4=73+5+13Q5=73+5+13+21
  1. Complete the expression below for Q6.

    Answer: Q6= -7 - 3 + + 13 + 21 +
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q6
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=7 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=73=7+S1

    and

    Q3=73+5=7+S2

    and so on. So the general term for Qn is

    Qn=7+Sn1

    Sn has a constant difference of d=8. So we can use this to find the missing values of Q6.

    Q6=73+5+13+21+29

    Submit your answer as: and
  2. Calculate the value of Q116.

    Answer: Q116=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q116 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=7+Sn1

    And we are given in the question that

    Sn=4n27n

    We can use these together to find the value of Q116.

    Q116=7+S115=7+4(115)27(115)=7+52900805=52088

    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first 30 terms of an arithmetic series are 4+7+10++91. Compute the sum of these 30 terms for the series, S30.

    Answer: S30=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first three terms and the last term of an arithmetic series, and we need to calculate the sum of the first 30 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 30 terms" as well as in the symbol "S30." Since we want to add up 30 terms of the series, n=30.

    The first term is given in the question: a=4. The last term is also given in the question l=91. (The question also shows the second and third terms, but we do not need them to answer the question.)

    We now have the values n=30,a=4 and l=91.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S30=302((4)+(91))=15(95)=1 425

    The sum of the first 30 terms for the series is S30=1425.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 29 terms of the series, S29 . Remember that the last term, given above, has a value of T30=91.

    Answer: S29= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S29. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S30. Think about what this means: it is the sum of the first 30 terms, as shown below.

    S30=T1+T2+T3++T29+T30

    Now we want to find the sum of the first 29 terms, which is the same as above except with the last term taken away! In other words, S29=S30T30. We can substitute the values we know into this equation to get the answer.

    S29=S30T30=(1425)(91)=1334

    The sum of the first 29 terms for the series is S29=1334.


    Submit your answer as:

Working with arithmetic series

  1. The difference between the fourth and sixth terms of an arithmetic series is 12 and the sum of the first 26 terms is 2028. Find the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 26 terms. Then write out another equation based on the difference between the sixth and fourth terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 26 terms
    [−1 point ⇒ 5 / 6 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 26 terms is 2028. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)2028=262(2a+(261)d)4056=26(2a+25d)156=2a+25d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 3 / 6 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "difference between the fourth and sixth terms ... is 12". We can write this as an equation: T6T4=12. Using the relationship Tn=a+(n1)d, we can write this equation in terms of a and d:

    T6T4=12(a+5d)(a+3d)=122d=12d=6

    This is fantastic: the a's cancelled so we could actually find the value of the common difference, d.


    STEP: Use the first equation and the value of d to find a
    [−1 point ⇒ 2 / 6 points left]

    We now have the two equations:

    2a+25d=156(1)d=6(2)

    Substitute d=6 into equation (1) to find the value of a.

    2a+25d=1562a+25(6)=1562a=6a=3

    We now have the value of the first term, T1=a=3, and the common difference, d=6.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 6 points left]

    Now use the values of a=3 and d=6 to calculate the values of T2 and T3.

    T2=a+d=3+(6)=9T3=a+2d=3+2(6)=15

    The first three terms of the sequence are: T1=3,T2=9,T3=15.


    Submit your answer as: andand
  2. Hence, if you are now told that S25 is equal to 1875, find the value of T26.

    Answer:

    T26=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S26 and S25.


    STEP: Compare the sums S26 and S25 to find the value of T26
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S26 and S25:

    S26=T1+T2+T3++T24+T25+T26S25=T1+T2+T3++T24+T25

    As you can see, the only difference between S26 and S25 is one term, T26. If we subtract T26 from S26 we will get S25. In other words, T26 is the difference between S26 and S25:

    T26=S26S25

    Now we can substitute in the values of the two sums (S26=2028 and S25=1875) and find the term we want:

    T26=S26S25=(2028)(1875)=153

    The correct answer is T26=153.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first three terms of an arithmetic series are 8;5;2. Determine the sum of the first 14 terms of the series, S14.

    Answer: S14=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−2 points ⇒ 2 / 4 points left]

    The question gives us the first three terms of an arithmetic series, and we need to calculate the sum of the first 14 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 14 terms" as well as in the symbol "S14." For this question, we want to add up 14 terms of the series, so n=14.

    The first term is given in the question: a=8. The common difference is not given so clearly in the question - we must calculate it by using the terms given: d=T2T1=(5)(8)=3.

    We now have the values n=14,a=8 and d=3.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 4 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S14=142(2(8)+(141)(3))=7((16)+(13)(3))=7((16)+39)=7(23)=161

    The sum of the first 14 terms for the series is S14=161.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 13 terms of the series, S13 , if the last term has a value of T14=31.

    Answer: S13= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T14=31 and need to determine the sum S13. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 14 terms, as shown below.

    S14=T1+T2+T3++T13+T14

    Now we want to find the sum of the first 13 terms, which is the same as above except with the last term taken away! In other words, S13=S14T14. We can substitute the values we know into this equation to get the answer.

    S13=S14T14=(161)(31)=130

    The sum of the first 13 terms for the series is S13=130.


    Submit your answer as:

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

8,5,2,,58,61
  1. Write down the sixth term (T6) of the sequence.

    Answer: T6=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the sixth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 3. So the sixth term can be calculated as:

    Tn=a+(n1)dT6=(8)+((6)1)(3)=7

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d61=(8)+(n1)(3)61=8+3n372=3nn=24

    Submit your answer as:
  3. Calculate the sum of all the positive numbers in the sequence.

    Answer: The sum of the positive numbers is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the positive terms
    [−3 points ⇒ 0 / 3 points left]

    The first three terms are negative, and then all of the rest of the terms are positive. So we need to exclude the first three terms when we determine the sum of the positive terms.

    So the positive terms of the sequence look like:

    1,4,,61

    There will be 21 terms in this sequence, since we are excluding three negative terms. And the new value of a will be 1. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S21=212[2(1)+((21)1)(3)]=21+661,531,5=651

    Submit your answer as:
  4. Consider the sequence

    8,5,2,,58,61,,2 869

    Determine the number of terms in this sequence that will be exactly divisible by 4.

    Answer: The number of terms divisible by 4 is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 4
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 8, which is divisible by 4.

    But the last term in the sequence is 2 869, which is not divisible by 4. So we need to work out the value of the last term which is divisible by 4.

    Since the common difference of the sequence is 3, we can count back from 2 869 in steps of 3 until we reach a number which is divisible by 4.

    TIP: Use your calculator if necessary to find the last value which is divisible by 4.
    2 8692 8662 8632 860

    So the last term which is divisible by 4 is 2 860.


    STEP: Continue the sequence to identify terms divisible by 4
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    8,5,2,1,4,7,10,13,16,,2 860

    The numbers divisible by 4 form a new sequence: 8,4,16,,2 860

    The new pattern is an arithmetic sequence with a=8 and d=12.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 4
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d2 860=8+(n1)(12)2 860=8+12n122 880=12n240=n

    Submit your answer as:

Evaluating expressions involving arithmetic series

Evaluate without using a calculator

8+9+10++3289+87+85+9
Answer:The expression is equal to:
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the quotient of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=98=1dB=8789=2

Now find the number of terms in each series:

For 8+9+10++32:Tn=a+(n1)d32=8+(nA1)(1)24=(nA1)(1)24=nA125=nAFor 89+87+85+9:Tn=a+(n1)d9=89+(nB1)(2)98=(nB1)(2)49=nB150=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the quotient from the question using the formula, Sn=n2(a+l).

SASB=8+9+10++3289+87+85+9=(252)(8+32)(502)(899)

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SASB=(252)(8+32)(502)(899)=(252)(40)25(80) Aha! We can cancel a factor of 40!=(252)25(12)=252(125)(12)=25100=14

Cool - we got the answer without having to do the more complicated calculations because there was an opportunity to cancel two large numbers (the 40's).

The final answer is 14.


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Using terms of an arithmetic sequence to find the sum

Consider two terms of an arithmetic sequence: the third and eighth terms are 35 and 70, respectively. What is the sum of the first 43 terms of the sequence?

Answer:S43=
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 6 / 6 points left]

We have two terms in an arithmetic sequence, T3 and T8, and we must find the sum of 43 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. Since the question does not tell us what the last term is, we will plan to use the second formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 3 / 6 points left]

Think about how the two terms given fit into the entire sum:

T1+T2+T3+T4+T5+T6+T7+T8+
T1+T2+35+T4+T5+T6+T7+70+

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T3 and T8: it is T8T3=7035=35. Now we use the formula Tn=a+(n1)d to do something clever:

T8T3=35(a+(81)d)(a+(31)d)=35Use the values of nfor the terms we have(a+7d)(a+2d)=35a+7da2d=355d=35d=355d=7

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 2 / 6 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T3=35.

Tn=a+(n1)dT3=a+(31)d35=a+(2)735=a+1421=a

STEP: Calculate the sum using Sn=n2(2a+(n1)d)
[−2 points ⇒ 0 / 6 points left]

Finally we have enough information to evaluate the sum the question asked for. As noted at the beginning, we will use the formula Sn=n2(2a+(n1)d).

Substitute in the values and evaluate:

Sn=n2(2a+(n1)d)S43=432(2(21)+(431)7)S43=432(42+(42)7)S43=432(336)S43=7224

The sum of the first 43 terms is 7224.


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Finding the number of terms in a series

  1. The following expression is an arithmetic series.

    38337837383

    Determine the number of terms in the series.

    Answer: The number of terms is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 3. We can then substitute into the formula as follows:

    3=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=383. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=378(383)=5.

    Substitute these values into the equation from above:

    3=383+(nlast1)(5)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 5 into the brackets.

    3=383+(nlast1)(5)3=383+5nlast53=388+5nlast385=5nlast77=nlast

    The n-value of the final term is 77.

    The calculation here shows that there are 77 terms in the series.


    Submit your answer as:
  2. Is the number 324 a term in the series. Justify your answer with a calculation.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 324 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 324 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 324 is a number in the series. In other words, if we write out every term in the series 38337837383, will we find 324 somewhere in the middle or will the numbers "jump" over it?

    We can work out the answer using the same method as the first question - by finding the value of n for the number 324. Basically, we will assume that 324 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    324=383+(n1)(5)324=383+5nlast5324=388+5n64=5n12,8=n

    This calculation shows that if the term is 324 then n=12,8.

    How does this calculation show us whether 324 is a term in the series or not? It depends on whether or not the value for n makes sense. In this case, we found that n12,8, which does not make sense because it is a decimal number (not a whole number). n represents the position of the term in the series. Finding a term in position n12,8 is the same as saying that you stood in a queue of 77 people and you were 12,8th person in the queue. That is impossible!

    The correct answer is no, the number is not a term in the series because the n-value is not a natural number.


    Submit your answer as:

Exercises

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=2+2+6+10+

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=2.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=2(2)=4

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(2)+(k1)(4)=2+4k4=4k6

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=4
    2. c=6

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=yx(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=nk=1(4k6)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(2)+(n1)(4)]=n2[4+4n4]=n2[4n8]=2n24n

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=4+4+12+20+

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=4.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=4(4)=8

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(4)+(k1)(8)=4+8k8=8k12

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=8
    2. c=12

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=yx(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=nk=1(8k12)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(4)+(n1)(8)]=n2[8+8n8]=n2[8n16]=4n28n

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3
Maths formulas

Consider the series:

Sn=5+3+11+19+

to n terms.

  1. We can find an expression for the general term of the series in the form Tk=bk+c. What are the values of b and c?

    Answer:
    1. b=
    2. c=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    First find the values of a and d for this series.


    STEP: Determine the values of a and d
    [−1 point ⇒ 1 / 2 points left]

    The formula for the general term of an arithmetic series is

    Tn=a+(n1)d

    a is the first term of the series. So here a=5.

    d is the constant difference between the terms of the series. We can determine this by finding the difference between any two consecutive terms of the series. Taking the difference of the first two terms:

    d=T2T1=3(5)=8

    STEP: Substitute into the formula and simplify
    [−1 point ⇒ 0 / 2 points left]

    To find Tk, we can now substitute the values of a and d into the formula for the general term and simplify.

    Tk=(5)+(k1)(8)=5+8k8=8k13

    We were looking for b and c in the expression Tk=bk+c. So the correct answers are

    1. b=8
    2. c=13

    Submit your answer as: and
  2. Write Sn in sigma notation, and then answer the two questions below about the expression you find. It should look something like

    Sn=yx(Tk)

    What symbols or expressions should be in the place of x and y above?

    Answer:
    1. x:
    2. y:
    equation
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise sigma notation in the Everything Maths textbook.


    STEP: Write Sn in sigma notation and read off the information you need
    [−2 points ⇒ 0 / 2 points left]

    Writing Sn in sigma notation, we get:

    Sn=nk=1(8k13)

    So the correct answers are

    1. x:k=1
    2. y:n

    Submit your answer as: and
  3. Derive an algebraic expression for Sn.

    INSTRUCTION: Simplify your answer completely.
    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute the values of a and d into the formula for the sum of the series.


    STEP: Substitute a and d into the formula for Sn
    [−3 points ⇒ 0 / 3 points left]

    The formula for the sum of n terms in an arithmetic series is:

    Sn=n2[2a+(n1)d]

    We need to substitute the values of a and d into this formula, and simplify.

    Sn=n2[2a+(n1)d]=n2[2(5)+(n1)(8)]=n2[10+8n8]=n2[8n18]=4n29n

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=6+2+10+18+

to n terms. Sn is also described by the expression

Sn=4n210n

Another sequence is defined as:

Q1=12Q2=126Q3=126+2Q4=126+2+10Q5=126+2+10+18
  1. Complete the expression below for Q7.

    Answer: Q7= 12 6 + 2 + 10 + 18 + +
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q7
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=12 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=126=12+S1

    and

    Q3=126+2=12+S2

    and so on. So the general term for Qn is

    Qn=12+Sn1

    Sn has a constant difference of d=8. So we can use this to find the missing values of Q7.

    Q7=126+2+10+18+26+34

    Submit your answer as: and
  2. Calculate the value of Q115.

    Answer: Q115=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q115 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=12+Sn1

    And we are given in the question that

    Sn=4n210n

    We can use these together to find the value of Q115.

    Q115=12+S114=12+4(114)210(114)=12+519841140=50832

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=42814

to n terms. Sn is also described by the expression

Sn=3n2+7n

Another sequence is defined as:

Q1=7Q2=7+4Q3=7+42Q4=7+428Q5=7+42814
  1. Complete the expression below for Q7.

    Answer: Q7= 7 + 4 2 8 14
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q7
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=7 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=7+4=7+S1

    and

    Q3=7+42=7+S2

    and so on. So the general term for Qn is

    Qn=7+Sn1

    Sn has a constant difference of d=6. So we can use this to find the missing values of Q7.

    Q7=7+428142026

    Submit your answer as: and
  2. Calculate the value of Q112.

    Answer: Q112=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q112 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=7+Sn1

    And we are given in the question that

    Sn=3n2+7n

    We can use these together to find the value of Q112.

    Q112=7+S111=73(111)2+7(111)=736963+777=36179

    Submit your answer as:

Arithmetic series

Adapted from DBE Nov 2015 Grade 12, P1, Q3.4
Maths formulas

A series is defined as

Sn=3159

to n terms. Sn is also described by the expression

Sn=2n2+5n

Another sequence is defined as:

Q1=7Q2=7+3Q3=7+31Q4=7+315Q5=7+3159
  1. Complete the expression below for Q7.

    Answer: Q7= 7 + 3 1 5 9
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]


    STEP: Fill in the missing values in Q7
    [−2 points ⇒ 0 / 2 points left]

    Taking a closer look at the terms of Q, we can see that each new term of Q is the same as the term before it, with one extra number at the end.

    Q1=7 is a random number. But each of the other numbers which are being added to Q is a term of Sn. For instance:

    Q2=7+3=7+S1

    and

    Q3=7+31=7+S2

    and so on. So the general term for Qn is

    Qn=7+Sn1

    Sn has a constant difference of d=4. So we can use this to find the missing values of Q7.

    Q7=7+31591317

    Submit your answer as: and
  2. Calculate the value of Q114.

    Answer: Q114=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the fact that you know Qn in terms of S, and that you are given an algebraic expression for Sn.


    STEP: Substitute the expression for Sn into Q114 and simplify
    [−3 points ⇒ 0 / 3 points left]

    From Question 1, we know that

    Qn=7+Sn1

    And we are given in the question that

    Sn=2n2+5n

    We can use these together to find the value of Q114.

    Q114=7+S113=72(113)2+5(113)=725538+565=24966

    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first 30 terms of an arithmetic series are 3+(5)+(7)++(61). Determine the sum of these 30 terms for the series, S30.

    Answer: S30=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first three terms and the last term of an arithmetic series, and we need to calculate the sum of the first 30 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 30 terms" as well as in the symbol "S30." Since we want to add up 30 terms of the series, n=30.

    The first term is given in the question: a=3. The last term is also given in the question l=61. (The question also shows the second and third terms, but we do not need them to answer the question.)

    We now have the values n=30,a=3 and l=61.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S30=302((3)+(61))=15(64)=960

    The sum of the first 30 terms for the series is S30=960.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 29 terms of the series, S29 . Remember that the last term, given above, has a value of T30=61.

    Answer: S29= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S29. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S30. Think about what this means: it is the sum of the first 30 terms, as shown below.

    S30=T1+T2+T3++T29+T30

    Now we want to find the sum of the first 29 terms, which is the same as above except with the last term taken away! In other words, S29=S30T30. We can substitute the values we know into this equation to get the answer.

    S29=S30T30=(960)(61)=899

    The sum of the first 29 terms for the series is S29=899.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 6 and the last term is 72. In total, there are 27 terms in the series. Compute the sum of these 27 terms, S27.

    Answer: S27=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first and last terms of an arithmetic series, and we need to calculate the sum of the first 27 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 27 terms" as well as in the symbol "S27." Since we want to add up 27 terms of the series, n=27.

    The first term is given in the question: a=6. We can also get the last term in the series directly from the question: l=72.

    We now have the values n=27,a=6 and l=72.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S27=272((6)+(72))=272(66)=891

    The sum of the first 27 terms for the series is S27=891.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 26 terms of the series, S26 . Remember that the last term, given above, has a value of T27=72.

    Answer: S26= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S26. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S27. Think about what this means: it is the sum of the first 27 terms, as shown below.

    S27=T1+T2+T3++T26+T27

    Now we want to find the sum of the first 26 terms, which is the same as above except with the last term taken away! In other words, S26=S27T27. We can substitute the values we know into this equation to get the answer.

    S26=S27T27=(891)(72)=819

    The sum of the first 26 terms for the series is S26=819.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first 21 terms of an arithmetic series are 3+(6)+(9)++(63). Determine the sum of these 21 terms for the series, S21.

    Answer: S21=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(a+l).


    STEP: Identify the values of n,a and l
    [−1 point ⇒ 2 / 3 points left]

    For this question we have the first three terms and the last term of an arithmetic series, and we need to calculate the sum of the first 21 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(a+l):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • l is the last term of the series

    We need the values of n,a and l - then we can substitute them into the formula and evaluate the answer.

    We can take the value of n straight from the question: it is in the words "the sum of these 21 terms" as well as in the symbol "S21." Since we want to add up 21 terms of the series, n=21.

    The first term is given in the question: a=3. The last term is also given in the question l=63. (The question also shows the second and third terms, but we do not need them to answer the question.)

    We now have the values n=21,a=3 and l=63.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(a+l)S21=212((3)+(63))=212(66)=693

    The sum of the first 21 terms for the series is S21=693.


    Submit your answer as:
  2. Based on the first question, find the sum of the first 20 terms of the series, S20 . Remember that the last term, given above, has a value of T21=63.

    Answer: S20= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We need to determine the sum S20. Since the question is only worth one mark, we can expect that there is a quick way (faster than the calculation in the first question).

    Compare this question to the first one: in the first question we found the sum S21. Think about what this means: it is the sum of the first 21 terms, as shown below.

    S21=T1+T2+T3++T20+T21

    Now we want to find the sum of the first 20 terms, which is the same as above except with the last term taken away! In other words, S20=S21T21. We can substitute the values we know into this equation to get the answer.

    S20=S21T21=(693)(63)=630

    The sum of the first 20 terms for the series is S20=630.


    Submit your answer as:

Working with arithmetic series

  1. The sum of the second and fifth terms of an arithmetic series is 33 and the sum of the first 37 terms is 2331. Find the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 7 / 7 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 37 terms. Then write out another equation based on the sum of the second and fifth terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 37 terms
    [−1 point ⇒ 6 / 7 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 37 terms is 2331. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)2331=372(2a+(371)d)4662=37(2a+36d)126=2a+36d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 4 / 7 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "sum of the second and fifth terms ... is 33". We can write this as an equation: T2+T5=33. In fact, we can use the relationship Tn=a+(n1)d to write the sum in terms of a and d:

    T2+T5=33(a+1d)+(a+4d)=332a+5d=33


    STEP: Solve the two equations simultaneously for a and d
    [−2 points ⇒ 2 / 7 points left]

    We now have the two equations:

    2a+36d=126(1)2a+5d=33(2)

    Use elimination to solve the two equations simultaneously (elimination is the best choice because both equations include 2a).

    2a+36d=126(1)2a+5d=33(2)Eqn (1) - (2):31d=93d=3And 2a+5(3)=332a=18a=9

    We now have the value of the first term, T1=a=9, and the common difference, d=3.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 7 points left]

    Now use the values of a=9 and d=3 to calculate the values of T2 and T3.

    T2=a+d=9+(3)=12T3=a+2d=9+2(3)=15

    The first three terms of the sequence are: T1=9,T2=12,T3=15.


    Submit your answer as: andand
  2. Hence, if you are now told that S38 is equal to 2451, find the value of T38.

    Answer:

    T38=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S38 and S37.


    STEP: Compare the sums S38 and S37 to find the value of T38
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S38 and S37:

    S38=T1+T2+T3++T36+T37+T38S37=T1+T2+T3++T36+T37

    As you can see, the only difference between S38 and S37 is one term, T38. If we subtract T38 from S38 we will get S37. In other words, T38 is the difference between S38 and S37:

    T38=S38S37

    Now we can substitute in the values of the two sums (S38=2451 and S37=2331) and find the term we want:

    T38=S38S37=(2451)(2331)=120

    The correct answer is T38=120.


    Submit your answer as:

Working with arithmetic series

  1. The sum of the fourth and fifth terms of an arithmetic series is -49 and the sum of the first 31 terms is -3968. Find the first three terms in the series.

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 7 / 7 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 31 terms. Then write out another equation based on the sum of the fourth and fifth terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 31 terms
    [−1 point ⇒ 6 / 7 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 31 terms is -3968. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)3968=312(2a+(311)d)7936=31(2a+30d)256=2a+30d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 4 / 7 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "sum of the fourth and fifth terms ... is -49". We can write this as an equation: T4+T5=49. In fact, we can use the relationship Tn=a+(n1)d to write the sum in terms of a and d:

    T4+T5=49(a+3d)+(a+4d)=492a+7d=49


    STEP: Solve the two equations simultaneously for a and d
    [−2 points ⇒ 2 / 7 points left]

    We now have the two equations:

    2a+30d=256(1)2a+7d=49(2)

    Use elimination to solve the two equations simultaneously (elimination is the best choice because both equations include 2a).

    2a+30d=256(1)2a+7d=49(2)Eqn (1) - (2):23d=207d=9And 2a+7(9)=492a=14a=7

    We now have the value of the first term, T1=a=7, and the common difference, d=9.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 7 points left]

    Now use the values of a=7 and d=9 to calculate the values of T2 and T3.

    T2=a+d=7+(9)=2T3=a+2d=7+2(9)=11

    The first three terms of the sequence are: T1=7,T2=2,T3=11.


    Submit your answer as: andand
  2. Hence, if you are now told that S30 is equal to -3705, find the value of T31.

    Answer:

    T31=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S31 and S30.


    STEP: Compare the sums S31 and S30 to find the value of T31
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S31 and S30:

    S31=T1+T2+T3++T29+T30+T31S30=T1+T2+T3++T29+T30

    As you can see, the only difference between S31 and S30 is one term, T31. If we subtract T31 from S31 we will get S30. In other words, T31 is the difference between S31 and S30:

    T31=S31S30

    Now we can substitute in the values of the two sums (S31=3968 and S30=3705) and find the term we want:

    T31=S31S30=(3968)(3705)=263

    The correct answer is T31=263.


    Submit your answer as:

Working with arithmetic series

  1. The difference between the third and sixth terms of an arithmetic series is 18 and the sum of the first 22 terms is 1210. What are the values of the first three terms?

    Answer:

    T1=

    T2=

    T3=

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    Use the equation Sn=n2(2a+(n1)d) to write an equation using the information about the sum of the first 22 terms. Then write out another equation based on the difference between the sixth and third terms.


    STEP: Use the formula Sn=n2(2a+(n1)d) to write an equation for the sum of the first 22 terms
    [−1 point ⇒ 5 / 6 points left]

    There is a lot of information in the question. We will start with the statement that the sum of the first 22 terms is 1210. This information fits with the formula Sn=n2(2a+(n1)d), which connects the sum of n terms of an arithmetic series to the values of a and d.

    Substitute in what we know to see what the equation can give us:

    Sn=n2(2a+(n1)d)1210=222(2a+(221)d)2420=22(2a+21d)110=2a+21d

    For the moment we are stuck because there are two variables in the equation (a and d), so we need to let this equation sit for the moment while we see what else we can get out of the question.


    STEP: Use the information about the other terms to write another equation
    [−2 points ⇒ 3 / 6 points left]

    In the work above, we got an equation with two variables, so the plan for the question is now clear: find another equation with the variables a and d so that we can solve the equations simultaneously. Let us try to do that using the information in the question about the other terms.

    In this case, the question states that the "difference between the third and sixth terms ... is 18". We can write this as an equation: T6T3=18. Using the relationship Tn=a+(n1)d, we can write this equation in terms of a and d:

    T6T3=18(a+5d)(a+2d)=183d=18d=6

    This is fantastic: the a's cancelled so we could actually find the value of the common difference, d.


    STEP: Use the first equation and the value of d to find a
    [−1 point ⇒ 2 / 6 points left]

    We now have the two equations:

    2a+21d=110(1)d=6(2)

    Substitute d=6 into equation (1) to find the value of a.

    2a+21d=1102a+21(6)=1102a=16a=8

    We now have the value of the first term, T1=a=8, and the common difference, d=6.


    STEP: Calculate the values of the second and third terms
    [−2 points ⇒ 0 / 6 points left]

    Now use the values of a=8 and d=6 to calculate the values of T2 and T3.

    T2=a+d=8+(6)=2T3=a+2d=8+2(6)=4

    The first three terms of the sequence are: T1=8,T2=2,T3=4.


    Submit your answer as: andand
  2. Hence, if you are now told that S21 is equal to 1092, find the value of T22.

    Answer:

    T22=

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can find the answer by comparing the sums S22 and S21.


    STEP: Compare the sums S22 and S21 to find the value of T22
    [−1 point ⇒ 0 / 1 points left]

    Sn is the sum of n terms of a series: Sn=T1+T2+T3++Tn. In this question we have S22 and S21:

    S22=T1+T2+T3++T20+T21+T22S21=T1+T2+T3++T20+T21

    As you can see, the only difference between S22 and S21 is one term, T22. If we subtract T22 from S22 we will get S21. In other words, T22 is the difference between S22 and S21:

    T22=S22S21

    Now we can substitute in the values of the two sums (S22=1210 and S21=1092) and find the term we want:

    T22=S22S21=(1210)(1092)=118

    The correct answer is T22=118.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 1 and the common difference is 3. Determine the sum of the first 13 terms, S13.

    Answer: S13=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−1 point ⇒ 2 / 3 points left]

    The question gives us the first and common difference of an arithmetic series, and we need to calculate the sum of the first 13 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 13 terms" as well as in the symbol "S13." For this question, we want to add up 13 terms of the series, so n=13.

    The first term is given in the question: a=1. The common difference is also clearly stated in the question: d=3.

    We now have the values n=13,a=1 and d=3.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S13=132(2(1)+(131)(3))=132(2+(12)(3))=132(2+36)=132(38)=247

    The sum of the first 13 terms for the series is S13=247.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 14 terms of the series, S14 , if the term following the last term is T14=40.

    Answer: S14= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T14=40 and need to determine the sum S14. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 13 terms, as shown below.

    S13=T1+T2+T3++T12+T13

    Now we want to find the sum of the first 14 terms, which is the same as the sum above, except with one more term added on! In other words, S14=S13+T14. We can substitute the values we know into this equation to get the answer.

    S14=S13+T14=(247)+(40)=287

    The sum of the first 14 terms for the series is S14=287.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first three terms of an arithmetic series are 7;10;13. Compute the sum of the first 15 terms of the series, S15.

    Answer: S15=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−2 points ⇒ 2 / 4 points left]

    The question gives us the first three terms of an arithmetic series, and we need to calculate the sum of the first 15 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 15 terms" as well as in the symbol "S15." For this question, we want to add up 15 terms of the series, so n=15.

    The first term is given in the question: a=7. The common difference is not given so clearly in the question - we must calculate it by using the terms given: d=T2T1=(10)(7)=3.

    We now have the values n=15,a=7 and d=3.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 4 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S15=152(2(7)+(151)(3))=152((14)+(14)(3))=152((14)+(42))=152(56)=420

    The sum of the first 15 terms for the series is S15=420.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 14 terms of the series, S14 , if the last term has a value of T15=49.

    Answer: S14= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T15=49 and need to determine the sum S14. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 15 terms, as shown below.

    S15=T1+T2+T3++T14+T15

    Now we want to find the sum of the first 14 terms, which is the same as above except with the last term taken away! In other words, S14=S15T15. We can substitute the values we know into this equation to get the answer.

    S14=S15T15=(420)(49)=371

    The sum of the first 14 terms for the series is S14=371.


    Submit your answer as:

Finding the sum for a finite arithmetic series

  1. The first term of an arithmetic series is 8 and the common difference is 2. Find the sum of the first 14 terms, S14.

    Answer: S14=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You can calculate the sum of the first n terms of an arithmetic series using the formula Sn=n2(2a+(n1)d).


    STEP: Identify the values of n,a and d
    [−1 point ⇒ 2 / 3 points left]

    The question gives us the first and common difference of an arithmetic series, and we need to calculate the sum of the first 14 terms in the series.

    The formula for the sum of terms in an arithmetic series is Sn=n2(2a+(n1)d):

    • Sn is the sum we want
    • n is the number of terms we are summing
    • a is the first term of the series
    • d is the common difference in the series

    We need the values of n,a and d - then we can substitute them into the formula and evaluate the answer.

    The value of n is sitting in the question: we can find it in the words "the sum of the first 14 terms" as well as in the symbol "S14." For this question, we want to add up 14 terms of the series, so n=14.

    The first term is given in the question: a=8. The common difference is also clearly stated in the question: d=2.

    We now have the values n=14,a=8 and d=2.


    STEP: Substitute into the formula and evaluate the answer
    [−2 points ⇒ 0 / 3 points left]

    Now substitute the values into the formula and work out the answer.

    Sn=n2(2a+(n1)d)S14=142(2(8)+(141)(2))=7(16+(13)(2))=7(16+(26))=7(10)=70

    The sum of the first 14 terms for the series is S14=70.


    Submit your answer as:
  2. Use the answer to the first question to find the sum of the first 15 terms of the series, S15 , if the term following the last term is T15=20.

    Answer: S15= 
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can solve this in the same way as the first question. However, there is a quicker way!


    STEP: Compare the new question to the first question to work out the answer
    [−1 point ⇒ 0 / 1 points left]

    We now have the new information T15=20 and need to determine the sum S15. One option is to use the formula again, as in the first question. However, there is a much quicker way (and the question is only worth one mark).

    Compare this question to the first one: in the first question we found the sum of the first 14 terms, as shown below.

    S14=T1+T2+T3++T13+T14

    Now we want to find the sum of the first 15 terms, which is the same as the sum above, except with one more term added on! In other words, S15=S14+T15. We can substitute the values we know into this equation to get the answer.

    S15=S14+T15=(70)+(20)=90

    The sum of the first 15 terms for the series is S15=90.


    Submit your answer as:

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

5,2,1,,76,79
  1. Write down the fifth term (T5) of the sequence.

    Answer: T5=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the fifth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 3. So the fifth term can be calculated as:

    Tn=a+(n1)dT5=(5)+((5)1)(3)=7

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d79=(5)+(n1)(3)79=53n+387=3nn=29

    Submit your answer as:
  3. Calculate the sum of all the negative numbers in the sequence.

    Answer: The sum of the negative numbers is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the negative terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are positive, and then all of the rest of the terms are negative. So we need to exclude the first two terms when we determine the sum of the negative terms.

    So the negative terms of the sequence look like:

    1,4,,79

    There will be 27 terms in this sequence, since we are excluding two positive terms. And the new value of a will be 1. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S27=272[2(1)+((27)1)(3)]=271 093,5+40,5=1080

    Submit your answer as:
  4. Consider the sequence

    5,2,1,,76,79,,3 742

    Determine the number of terms in this sequence that will be exactly divisible by 5.

    Answer: The number of terms divisible by 5 is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 5, which is divisible by 5.

    But the last term in the sequence is 3 742, which is not divisible by 5. So we need to work out the value of the last term which is divisible by 5.

    Since the common difference of the sequence is 3, we can count back from 3 742 in steps of 3 until we reach a number which is divisible by 5.

    TIP: Use your calculator if necessary to find the last value which is divisible by 5.
    3 7423 7393 7363 7333 730

    So the last term which is divisible by 5 is 3 730.


    STEP: Continue the sequence to identify terms divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    5,2,1,4,7,10,13,16,19,22,25,,3 730

    The numbers divisible by 5 form a new sequence: 5,10,25,,3 730

    The new pattern is an arithmetic sequence with a=5 and d=15.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 5
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d3 730=5+(n1)(15)3 730=515n+153 750=15n250=n

    Submit your answer as:

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

6,2,2,,82,86
  1. Write down the fifth term (T5) of the sequence.

    Answer: T5=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the fifth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 4. So the fifth term can be calculated as:

    Tn=a+(n1)dT5=(6)+((5)1)(4)=10

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d86=(6)+(n1)(4)86=64n+496=4nn=24

    Submit your answer as:
  3. Calculate the sum of all the negative numbers in the sequence.

    Answer: The sum of the negative numbers is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the negative terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are positive, and then all of the rest of the terms are negative. So we need to exclude the first two terms when we determine the sum of the negative terms.

    So the negative terms of the sequence look like:

    2,6,,86

    There will be 22 terms in this sequence, since we are excluding two positive terms. And the new value of a will be 2. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S22=222[2(2)+((22)1)(4)]=44968,0+44,0=968

    Submit your answer as:
  4. Consider the sequence

    6,2,2,,82,86,,2 630

    Determine the number of terms in this sequence that will be exactly divisible by 3.

    Answer: The number of terms divisible by 3 is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 3
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 6, which is divisible by 3.

    But the last term in the sequence is 2 630, which is not divisible by 3. So we need to work out the value of the last term which is divisible by 3.

    Since the common difference of the sequence is 4, we can count back from 2 630 in steps of 4 until we reach a number which is divisible by 3.

    TIP: Use your calculator if necessary to find the last value which is divisible by 3.
    2 6302 6262 622

    So the last term which is divisible by 3 is 2 622.


    STEP: Continue the sequence to identify terms divisible by 3
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    6,2,2,6,10,14,18,,2 622

    The numbers divisible by 3 form a new sequence: 6,6,18,,2 622

    The new pattern is an arithmetic sequence with a=6 and d=12.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 3
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d2 622=6+(n1)(12)2 622=612n+122 640=12n220=n

    Submit your answer as:

Finite arithmetic sequence

Adapted from DBE Nov 2016 Grade 12, P1, Q2
Maths formulas

Given the finite arithmetic sequence:

5,1,3,,99,103
  1. Write down the fifth term (T5) of the sequence.

    Answer: T5=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Determine the value of the fifth term
    [−1 point ⇒ 0 / 1 points left]

    We can see from the first three terms that the common difference value for this sequence is 4. So the fifth term can be calculated as:

    Tn=a+(n1)dT5=(5)+((5)1)(4)=11

    Submit your answer as:
  2. Calculate the number of terms in the sequence.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise arithmetic sequences in the Everything Maths textbook.


    STEP: Calculate the number of terms in the sequence
    [−3 points ⇒ 0 / 3 points left]

    We can use the same formula that we used in Question 1 to find the number of terms in the sequence, because we know the value of the final term in the sequence:

    Tn=a+(n1)d103=(5)+(n1)(4)103=5+4n4112=4nn=28

    Submit your answer as:
  3. Calculate the sum of all the positive numbers in the sequence.

    Answer: The sum of the positive numbers is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Calculate the sum of the positive terms
    [−3 points ⇒ 0 / 3 points left]

    The first two terms are negative, and then all of the rest of the terms are positive. So we need to exclude the first two terms when we determine the sum of the positive terms.

    So the positive terms of the sequence look like:

    3,7,,103

    There will be 26 terms in this sequence, since we are excluding two negative terms. And the new value of a will be 3. We can then calculate the sum of these terms using the formula for the sum of an arithmetic series:

    Sn=n2[2a+(n1)d]S26=262[2(3)+((26)1)(4)]=78+1 352,052,0=1378

    Submit your answer as:
  4. Consider the sequence

    5,1,3,,99,103,,4 991

    Determine the number of terms in this sequence that will be exactly divisible by 5.

    Answer: The number of terms divisible by 5 is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite arithmetic series in the Everything Maths textbook.


    STEP: Determine the last term in the sequence divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The first term in the sequence is 5, which is divisible by 5.

    But the last term in the sequence is 4 991, which is not divisible by 5. So we need to work out the value of the last term which is divisible by 5.

    Since the common difference of the sequence is 4, we can count back from 4 991 in steps of 4 until we reach a number which is divisible by 5.

    TIP: Use your calculator if necessary to find the last value which is divisible by 5.
    4 9914 9874 9834 9794 975

    So the last term which is divisible by 5 is 4 975.


    STEP: Continue the sequence to identify terms divisible by 5
    [−0 points ⇒ 4 / 4 points left]

    The original sequence is:

    5,1,3,7,11,15,19,23,27,31,35,,4 975

    The numbers divisible by 5 form a new sequence: 5,15,35,,4 975

    The new pattern is an arithmetic sequence with a=5 and d=20.


    STEP: Use the formula for arithmetic sequences to calculate the number of terms divisible by 5
    [−4 points ⇒ 0 / 4 points left]
    Tn=a+(n1)d4 975=5+(n1)(20)4 975=5+20n205 000=20n250=n

    Submit your answer as:

Evaluating expressions involving arithmetic series

Evaluate the following difference without using a calculator:

(137471)(214472)
Answer:The expression is equal to:
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the difference of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=31=4dB=12=3

Now find the number of terms in each series:

For 137471:Tn=a+(n1)d471=1+(nA1)(4)472=(nA1)(4)118=nA1119=nAFor 214472:Tn=a+(n1)d472=2+(nB1)(3)474=(nB1)(3)158=nB1159=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the difference from the question using the formula, Sn=n2(a+l).

SASB=(137471)(214472)=(1192(1471))(1592(2472))

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SASB=(1192(1471))(1592(2472))=1192(470)1592(470) Check this: we can factorise the -470's out!=(470)(11921592)=(470)(1191592)=(470)(402)=(470)(20)=9 400

By using factorisation, we were able to work out the question without the more complex calculations.

The final answer is 9 400.


Submit your answer as:

Evaluating expressions involving arithmetic series

Evaluate without using a calculator

(8+12+16++472)(12+18+24++468)
Answer:The expression is equal to:
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the difference of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=128=4dB=1812=6

Now find the number of terms in each series:

For 8+12+16++472:Tn=a+(n1)d472=8+(nA1)(4)464=(nA1)(4)116=nA1117=nAFor 12+18+24++468:Tn=a+(n1)d468=12+(nB1)(6)456=(nB1)(6)76=nB177=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the difference from the question using the formula, Sn=n2(a+l).

SASB=(8+12+16++472)(12+18+24++468)=(1172(8+472))(772(12+468))

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SASB=(1172(8+472))(772(12+468))=1172(480)772(480) Check this: we can factorise the 480's out!=(480)(1172772)=(480)(117772)=(480)(402)=(480)(20)=9 600

By using factorisation, we were able to work out the question without the more complex calculations.

The final answer is 9 600.


Submit your answer as:

Evaluating expressions involving arithmetic series

Evaluate the following expression without using a calculator:

(10+6+2210)(517205)
Answer:The expression is equal to:
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The question states that you should not use a calculator. That is a big hint that the expression can be simplified somehow so that the question becomes much easier than it looks.


STEP: Examine the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

The expression in the question is complex. However, the question says we must find the answer "without using a calculator". This means two things: (a) we must show all of the steps to get full marks and (b) there is probably a way to simplify the expression without doing lots of difficult calculations.

The first thing you hopefully noticed is that the expression is the difference of two different arithmetic series. The plan will be to use the formula Sn=n2(a+l) to rewrite both of the series: that will make the expression simpler. Then we will see what type of calculations we must do.


STEP: Find the number of terms in each series
[−2 points ⇒ 3 / 5 points left]

In order to use the formula Sn=n2(a+l), we need the values for n, a and l for each of the series. For both series, a and l are sitting in the question. However, the number of terms, n, is not obvious. We need to calculate the number of terms for both series. For this we can use the last term in each sequence in the formula Tn=a+(n1)d.

We must find the common difference (d) for each of the series first:

dA=610=4dB=15=6

Now find the number of terms in each series:

For 10+6+2210:Tn=a+(n1)d210=10+(nA1)(4)220=(nA1)(4)55=nA156=nAFor 517205:Tn=a+(n1)d205=5+(nB1)(6)210=(nB1)(6)35=nB136=nB

STEP: Use the sum formula Sn=n2(a+l) to rewrite the expression in the question
[−1 point ⇒ 2 / 5 points left]

Now use these values, together with a and l for each series, to rewrite the difference from the question using the formula, Sn=n2(a+l).

SASB=(10+6+2210)(517205)=(562(10210))(362(5205))

STEP: Simplify the expression (remember: no calculator!)
[−2 points ⇒ 0 / 5 points left]

From this point, we should start to simplify the expression. However, remember that the question says that we should evaluate the expression without a calculator. That means that somewhere during the calculations, there will probably be an opportunity to simplify the expression in some clever way which makes the calculation much easier... so keep your eyes open!

SASB=(562(10210))(362(5205))=28(200)18(200) Check this: we can factorise the -200's out!=(200)(2818)=(200)(10)=2 000

By using factorisation, we were able to work out the question without the more complex calculations.

The final answer is 2 000.


Submit your answer as:

Using terms of an arithmetic sequence to find the sum

Consider two terms of an arithmetic sequence: the forty-eighth and fiftieth terms are 284 and 296, respectively. What is the sum of the first 50 terms of the sequence?

Answer:S50=
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

We have two terms in an arithmetic sequence, T48 and T50, and we must find the sum of 50 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. This question tells us the value of the last term, T50, so we will use the first formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 2 / 5 points left]

Think about how the two terms given fit into the entire sum:

+T47+T48+T49+T50
+T47+284+T49+296

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T48 and T50: it is T50T48=296284=12. Now we use the formula Tn=a+(n1)d to do something clever:

T50T48=12(a+(501)d)(a+(481)d)=12Use the values of nfor the terms we have(a+49d)(a+47d)=12a+49da47d=122d=12d=122d=6

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 1 / 5 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T48=284.

Tn=a+(n1)dT48=a+(481)d284=a+(47)6284=a+2822=a

STEP: Calculate the sum using Sn=n2(a+l)
[−1 point ⇒ 0 / 5 points left]

Finally we have enough information to evaluate the sum the question asked for. As mentioned at the beginning, we can use the formula Sn=n2(a+l) because the question gives us the last term, T50. (We could also use the other formula, Sn=n2(2a+(n1)d), since we know that d=6; we would get the same answer, but the first formula is simpler so it is a wiser choice.)

Substitute in the values and evaluate:

Sn=n2(a+l)S50=502(2+296)S50=25(298)S50=7450

The sum of the first 50 terms is 7450.


Submit your answer as:

Using terms of an arithmetic sequence to find the sum

Consider two terms of an arithmetic sequence: the twentieth and twenty-fourth terms are -151 and -191, respectively. Find the sum of the first 24 terms of the sequence.

Answer:S24=
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 5 / 5 points left]

We have two terms in an arithmetic sequence, T20 and T24, and we must find the sum of 24 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. This question tells us the value of the last term, T24, so we will use the first formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 2 / 5 points left]

Think about how the two terms given fit into the entire sum:

+T19+T20+T21+T22+T23+T24
+T19+(151)+T21+T22+T23+(191)

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T20 and T24: it is T24T20=(191)(151)=40. Now we use the formula Tn=a+(n1)d to do something clever:

T24T20=40(a+(241)d)(a+(201)d)=40Use the values of nfor the terms we have(a+23d)(a+19d)=40a+23da19d=404d=40d=404d=10

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 1 / 5 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T20=151.

Tn=a+(n1)dT20=a+(201)d151=a+(19)(10)151=a19039=a

STEP: Calculate the sum using Sn=n2(a+l)
[−1 point ⇒ 0 / 5 points left]

Finally we have enough information to evaluate the sum the question asked for. As mentioned at the beginning, we can use the formula Sn=n2(a+l) because the question gives us the last term, T24. (We could also use the other formula, Sn=n2(2a+(n1)d), since we know that d=10; we would get the same answer, but the first formula is simpler so it is a wiser choice.)

Substitute in the values and evaluate:

Sn=n2(a+l)S24=242(39+(191))S24=12(152)S24=1824

The sum of the first 24 terms is -1824.


Submit your answer as:

Using terms of an arithmetic sequence to find the sum

The third term of an arithmetic sequence is 26 and the fifth term is 42. What is the sum of the first 37 terms of the sequence?

Answer:S37=
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by using the two terms given in the question to find the common difference and then find the first term. After you have those numbers you can use one of the formulae for the sum of an arithmetic series.


STEP: Think about the information given in the question and consider how to solve the problem
[−0 points ⇒ 6 / 6 points left]

We have two terms in an arithmetic sequence, T3 and T5, and we must find the sum of 37 terms for the sequence. There are two formulae for the sum of an arithmetic sequence: Sn=n2(a+l) and Sn=n2(2a+(n1)d). You can see that we either need to find a and l to use the first version of the formula, or a and d to use the second version. Since the question does not tell us what the last term is, we will plan to use the second formula.


STEP: Use the two terms from the question to find the common difference of the sequence
[−3 points ⇒ 3 / 6 points left]

Think about how the two terms given fit into the entire sum:

T1+T2+T3+T4+T5+
T1+T2+26+T4+42+

This list can help you see that if we have any two terms from an arithmetic sequence, we can always use them to find the common difference, d, (because each term is always separated by d). Specifically, the difference between any two terms is related to the common difference.

Let us find the difference between T3 and T5: it is T5T3=4226=16. Now we use the formula Tn=a+(n1)d to do something clever:

T5T3=16(a+(51)d)(a+(31)d)=16Use the values of nfor the terms we have(a+4d)(a+2d)=16a+4da2d=162d=16d=162d=8

STEP: Use the common difference and the formula Tn=a+(n1)d to find the first term
[−1 point ⇒ 2 / 6 points left]

Now that we know the value of d, we can find the value of a. Do this using the formula Tn=a+(n1)d with either of the terms given in the question. We will use the first term, T3=26.

Tn=a+(n1)dT3=a+(31)d26=a+(2)826=a+1610=a

STEP: Calculate the sum using Sn=n2(2a+(n1)d)
[−2 points ⇒ 0 / 6 points left]

Finally we have enough information to evaluate the sum the question asked for. As noted at the beginning, we will use the formula Sn=n2(2a+(n1)d).

Substitute in the values and evaluate:

Sn=n2(2a+(n1)d)S37=372(2(10)+(371)8)S37=372(20+(36)8)S37=372(308)S37=5698

The sum of the first 37 terms is 5698.


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Finding the number of terms in a series

  1. The following expression shows an arithmetic series.

    20+19+18+34

    How many terms are there in the series?

    Answer: The number of terms is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 4. We can then substitute into the formula as follows:

    4=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=20. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=1920=1.

    Substitute these values into the equation from above:

    4=20+(nlast1)(1)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 1 into the brackets.

    4=20+(nlast1)(1)4=20nlast+14=21nlastnlast=25(rearrange to makenlast positive)

    The n-value of the final term is 25.

    The calculation here shows that there are 25 terms in the series.


    Submit your answer as:
  2. Is the number 4 a term in the series. Justify your answer with a calculation.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 4 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 4 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 4 is a number in the series. In other words, if we write out every term in the series 20+19+18+34, will we find 4 somewhere in the middle or will the numbers "jump" over it?

    For the series 20+19+18+34, it is clear that the number 4 is in the series. However, we will show that it is true with a calculation, as the question instructs.

    We can work out the answer using the same method as the first question - by finding the value of n for the number 4. Basically, we will assume that 4 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    4=20+(n1)(1)4=20nlast+14=21nn=17(rearrange to maken positive)

    This calculation shows that if the term is 4 then n=17.

    How does this calculation show us whether 4 is a term in the series or not? It depends on whether or not the value for n makes sense. The answer n=17 does make sense because it is a natural number between 1 and 25. n=17 means that 4 is the 17th term in the series. (For comparison, suppose we got the answer n=17 or n=26; those answers are impossible, because we already know that the series starts at the first term, n=1, and ends at n=25.)

    The correct answer is yes, the number is a term in the series because the n-value is a natural number.


    Submit your answer as:

Finding the number of terms in a series

  1. The following expression is an arithmetic series.

    7+31325329

    Determine the number of terms in the series.

    Answer: The number of terms is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 329. We can then substitute into the formula as follows:

    329=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=7. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=37=4.

    Substitute these values into the equation from above:

    329=7+(nlast1)(4)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 4 into the brackets.

    329=7+(nlast1)(4)329=74nlast+4329=114nlast4nlast=340(rearrange to makenlast positive)nlast=85

    The n-value of the final term is 85.

    The calculation here shows that there are 85 terms in the series.


    Submit your answer as:
  2. Is the number 191 a term in the series. Use a calculation to justify your answer.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 191 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 191 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 191 is a number in the series. In other words, if we write out every term in the series 7+31325329, will we find 191 somewhere in the middle or will the numbers "jump" over it?

    We can work out the answer using the same method as the first question - by finding the value of n for the number 191. Basically, we will assume that 191 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    191=7+(n1)(4)191=74nlast+4191=114n4n=202(rearrange to maken positive)n=50,5

    This calculation shows that if the term is 191 then n=50,5.

    How does this calculation show us whether 191 is a term in the series or not? It depends on whether or not the value for n makes sense. In this case, we found that n50,5, which does not make sense because it is a decimal number (not a whole number). n represents the position of the term in the series. Finding a term in position n50,5 is the same as saying that you stood in a queue of 85 people and you were 50,5th person in the queue. That is impossible!

    The correct answer is no, the number is not a term in the series because the n-value is not a natural number.


    Submit your answer as:

Finding the number of terms in a series

  1. The following expression is an arithmetic series.

    2+4+10++208+214

    Determine the number of terms in the series.

    Answer: The number of terms is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by finding the values of a and d. Then use the formula Tn=a+(n1)d.


    STEP: Set up the equation using the last term
    [−1 point ⇒ 2 / 3 points left]

    The question gives us an arithmetic series and we must determine the number of terms in the series. We can do this using the formula Tn=a+(n1)d.

    The number of terms in the series is represented by the n-value of the last term in the series. In this case, the last term is 214. We can then substitute into the formula as follows:

    214=a+(nlast1)d

    If we also substitute in the values of the first term, a, and the common difference, d, we can solve the equation for the value of n for the last term. This will give us the total number of terms in the series.


    STEP: Find the values of a and d and substitute into the equation
    [−1 point ⇒ 1 / 3 points left]

    From the series in the question, we can get both the first term and the common difference. The first term is sitting there for us to take: a=2. The common difference requires a bit of work: it is the difference between any two consecutive terms in the series: d=T2T1, so we get d=4(2)=6.

    Substitute these values into the equation from above:

    214=2+(nlast1)(6)

    STEP: Solve the equation for n
    [−1 point ⇒ 0 / 3 points left]

    Now we can solve the equation for nlast. Start by distributing the 6 into the brackets.

    214=2+(nlast1)(6)214=2+6nlast6214=8+6nlast222=6nlast37=nlast

    The n-value of the final term is 37.

    The calculation here shows that there are 37 terms in the series.


    Submit your answer as:
  2. Determine if 99 is one of the terms in the series or not. Justify your answer with a calculation.

    NOTE: You will pick the correct choice below, but on a test or exam, you must show the calculation to get full marks.
    Answer:

    Is 99 in the series?

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You can use the same method as in the first question to find the answer: substitute values in the equation Tn=a+(n1)d and find the value of n. The answer you get for n will tell you if 99 belongs in the series or not.


    STEP: Use the formula Tn=a+(n1)d to find the value of n
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to figure out whether or not 99 is a number in the series. In other words, if we write out every term in the series 2+4+10++208+214, will we find 99 somewhere in the middle or will the numbers "jump" over it?

    We can work out the answer using the same method as the first question - by finding the value of n for the number 99. Basically, we will assume that 99 is a term in the series and then use the value of n to check whether or not it is true! Substitute the values into the equation and calculate n.

    99=2+(n1)(6)99=2+6nlast699=8+6n107=6n17,8333...=n17,8n

    This calculation shows that if the term is 99 then n17,8.

    How does this calculation show us whether 99 is a term in the series or not? It depends on whether or not the value for n makes sense. In this case, we found that n17,8, which does not make sense because it is a decimal number (not a whole number). n represents the position of the term in the series. Finding a term in position n17,8 is the same as saying that you stood in a queue of 37 people and you were 17,8th person in the queue. That is impossible!

    The correct answer is no, the number is not a term in the series because the n-value is not a natural number.


    Submit your answer as:

4. Geometric sequences

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

18;9;4,5;2,25;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=18+(12)(n1)
B Tn=18(12)n1
C Tn=18(12)n
D Tn=12(18)n
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=18.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=918=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=18(12)n1

So the correct choice is Option B.


Submit your answer as:

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    2;4;8;
    1. Determine the constant ratio, r.
    2. Calculate the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 2;4;8; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the first and second terms:

    r=T2T1r=42=2

    Notice that the ratio is negative: that is the reason why the terms in the sequence keep changing signs. (Multiplying by a negative always changes the sign.)


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 2;4;8;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 2;4;8. So, the value for a is 2. Substituting a=2 and r=2 into the general formula, we get:

    Tn=arn1=(2)(2)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 2 and 2. Start by simplifying the exponential part of the expression, (2)n1.

    Tn=(2)(2)n1=(2)(2)n(2)1=(2)(2)n(12)=(2)n

    Therefore for the given sequence, the general formula can be written as Tn=(2)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    5;15x2;45x4;
    1. Determine the constant ratio, r.
    2. Calculate the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 5;15x2;45x4;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T3T2r=45x415x2r=3x2

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=5 while the constant ratio is r=3x2. Substitute these values into the formula:

    Tn=arn1=(5)(3x2)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (3x2)n1 using the exponent law.

    Tn=(5)(3x2)n1=(5)(3x2)n(3x2)1=(5)(13x2)(3x2)n=(53x2)(3x2)n

    Therefore for the given sequence, the general formula can be written as Tn=(53x2)(3x2)n.


    Submit your answer as: and

The geometric mean

  1. Determine the value of p if the following three terms are the beginning of a geometric sequence:

    12;p;2,if p<0

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: p=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The number p is the geometric mean of 12 and 2. Use the constant ratio to write an equation which includes p. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the unknown term in the geometric sequence 12;p;2.

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two successive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2p(12)=2p

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for p.

    p(12)=2pp2=(2)(12)p2=1p=±1Don't forget the plus-minuswhich comes with the square root!p=±1

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 12;1;2 while if the constant ratio is negative the sequence will be 12;1;2.

    NOTE:

    While it does not say so, this question is about the geometric mean. If we have three terms in a geometric sequence, the term in the middle is the geometric mean of the other two. In this question, p is the geometric mean of 12 and 2, and it has two possible values: ±1.

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is negative. So we can ignore the 1 answer.

    The correct answer is 1.


    Submit your answer as:
  2. Determine the common ratio for the three terms in Question 1.

    INSTRUCTION:If there is more than one answer, separate the answers by the ';' symbol.
    Answer: Common ratio =
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The common ratio is the quotient of any two consecutive terms in the sequence.
    STEP: Use two consecutive terms to calculate r
    [−3 points ⇒ 0 / 3 points left]

    We need to find the common ratio for the geometric sequence. We know from Question 1 that the sequence is 12;1;2.

    The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can use any two consecutive terms in the sequence to find the answer:

    r=T3T2orr=T2T1

    We can summarise this relationship as r=TnTn1. In this solution we will use the second and third terms.

    r=T3T2r=2(1)r=2

    Now we can see why the terms in the sequence alternate signs: the constant ratio is negative.

    The constant ratio is r=2.


    Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

18;9;4,5;2,25;

Calculate the value of the fifth term, T5, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T5=
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the fifth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=18.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=918=12

STEP: Calculate the value of the fifth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the fifth term.

T5=(18)(12)(51)=(18)(116)=1,125

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Exercises

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

10;5;2,5;1,25;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=10(5)n1
B Tn=10(12)n1
C Tn=10+(12)(n1)
D Tn=12(10)n1
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=10.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=510=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=10(12)n1

So the correct choice is Option B.


Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

14;7;3,5;1,75;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=12(14)n1
B Tn=14(12)n1
C Tn=14+(12)(n1)
D Tn=14(7)n1
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=14.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=714=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=14(12)n1

So the correct choice is Option B.


Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.2
Maths formulas

The following geometric sequence is given:

20;10;5;2,5;

Which of the following is the correct expression for the nth term of this sequence?

A Tn=20+(12)(n1)
B Tn=20(12)n1
C Tn=20(10)n1
D Tn=20(2)n
Answer: The correct choice is Option .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

First determine the values of the first term and the constant ratio of the sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

We need to know the values of a and r in order to identify the correct equation.

a is the first term of the sequence, so here a=20.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=1020=12

STEP: Substitute a and r into the general formula
[−1 point ⇒ 0 / 2 points left]

Substitute these values into the formula for the general term of a geometric sequence, so we can identify the correct expression:

Tn=arn1=20(12)n1

So the correct choice is Option B.


Submit your answer as:

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    13;19;127;
    1. Find the constant ratio, r.
    2. Determine the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 13;19;127; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the first and second terms:

    r=T2T1r=19÷13r=19×3=13


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 13;19;127;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 13;19;127. So, the value for a is 13. Substituting a=13 and r=13 into the general formula, we get:

    Tn=arn1=(13)(13)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 13 and 13. Start by simplifying the exponential part of the expression, 13n1.

    Tn=(13)(13)n1=(13)(13)n(13)1=(13)(13)n(113)=(13)n

    Therefore for the given sequence, the general formula can be written as Tn=(13)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    1;x;x2;
    1. Find the constant ratio, r.
    2. Determine the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 1;x;x2;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T3T2r=x2xr=x

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=1 while the constant ratio is r=x. Substitute these values into the formula:

    Tn=arn1=(1)(x)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (x)n1 using the exponent law.

    Tn=(1)(x)n1=(1)(x)n(x)1=(1)(1x)(x)n=(1x)(x)n

    Therefore for the given sequence, the general formula can be written as Tn=(1x)(x)n.


    Submit your answer as: and

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    12;14;18;
    1. Determine the constant ratio, r.
    2. Find the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 12;14;18; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the second and third terms:

    r=T3T2r=18÷14r=18×4=12

    Notice that the ratio is negative: that is the reason why the terms in the sequence keep changing signs. (Multiplying by a negative always changes the sign.)


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 12;14;18;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 12;14;18. So, the value for a is 12. Substituting a=12 and r=12 into the general formula, we get:

    Tn=arn1=(12)(12)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 12 and 12. Start by simplifying the exponential part of the expression, (12)n1.

    Tn=(12)(12)n1=(12)(12)n(12)1=(12)(12)n(112)=(12)n

    Therefore for the given sequence, the general formula can be written as Tn=(12)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    4x;16x2;64x3;
    1. Determine the constant ratio, r.
    2. Find the general formula for the nth term of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 4x;16x2;64x3;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T3T2r=64x316x2r=4x

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=4x while the constant ratio is r=4x. Substitute these values into the formula:

    Tn=arn1=(4x)(4x)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (4x)n1 using the exponent law.

    Tn=(4x)(4x)n1=(4x)(4x)n(4x)1=(4x)(14x)(4x)n=(1)(4x)n

    Therefore for the given sequence, the general formula can be written as Tn=(1)(4x)n.


    Submit your answer as: and

Constant ratio and general formula

  1. The first three terms of a geometric sequence are given:

    12;14;18;
    1. Determine the constant ratio, r.
    2. Determine the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    numeric
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The constant ratio, r, is the quotient of one term and the term before it. For example,

    r=T3T2

    STEP: Calculate the value of r
    [−1 point ⇒ 1 / 2 points left]

    We have the geometric sequence 12;14;18; and we need to figure out the constant ratio as well as the general formula. The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can calculate the ratio by finding the quotient of any two consecutive terms: r=T3T2 or T2T1 or TnTn1. Therefore, we can pick any pair of consecutive terms from the sequence. This time let's use the first and second terms:

    r=T2T1r=14÷12r=14×2=12


    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 2 points left]

    Now that we have calculated the value for r, we can calculate the general formula for 12;14;18;. To find the general formula for a geometric sequence, we use the equation Tn=arn1. The variable a is the first term in the sequence and r is the constant ratio.

    The question states that the first three terms of the sequence are 12;14;18. So, the value for a is 12. Substituting a=12 and r=12 into the general formula, we get:

    Tn=arn1=(12)(12)n1
    NOTE:

    While it is not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Remember that we calculate exponents before multiplying (according to BODMAS) so do not multiply 12 and 12. Start by simplifying the exponential part of the expression, 12n1.

    Tn=(12)(12)n1=(12)(12)n(12)1=(12)(12)n(112)=(12)n

    Therefore for the given sequence, the general formula can be written as Tn=(12)n


    Submit your answer as: and
  2. The first three terms of a geometric sequence are given:

    2x;4x2;8x3;
    1. Determine the constant ratio, r.
    2. Determine the formula for Tn of the sequence.
    INSTRUCTION: Give your answer as an expression in terms of x and n, and without any decimals. Other simplification is not necessary for the purposes of this question.
    Answer:
    1. r=
    2. Tn=
    expression
    one-of
    type(expression)
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do not let the variables in the sequence bother you: this question can be solved in the same way as the first question.


    STEP: Calcuate the value of r
    [−2 points ⇒ 1 / 3 points left]

    As with the first question, we have a geometric sequence, 2x;4x2;8x3;, and we need to figure out the constant ratio as well as the general formula. We calculate the ratio for this sequence in the same way as for the previous question: use any two consecutive terms: r=T3T2 or T2T1 or Tn1Tn. Since we can choose which terms we use, pick the simplest ones available for the calculation. Then simplify.

    r=T3T2r=8x34x2r=2x

    STEP: Substitute a and r into the general formula
    [−1 point ⇒ 0 / 3 points left]

    Now we will find the general formula for the sequence. This follows the exact same process as in the first question: we use Tn=arn1 and substitute in the correct values for a and r for this sequence. The first term of the sequence is a=2x while the constant ratio is r=2x. Substitute these values into the formula:

    Tn=arn1=(2x)(2x)n1
    NOTE:

    While it is also not necessary for this question, we could express Tn with an exponent of n instead of n1.

    Start by simplifying (2x)n1 using the exponent law.

    Tn=(2x)(2x)n1=(2x)(2x)n(2x)1=(2x)(12x)(2x)n=(1)(2x)n

    Therefore for the given sequence, the general formula can be written as Tn=(1)(2x)n.


    Submit your answer as: and

The geometric mean

  1. You are given the terms

    14;y;4

    which form a geometric sequence. What is the value of y if y>0? Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: y=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The number y is the geometric mean of 14 and 4. Use the constant ratio to write an equation which includes y. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the unknown term in the geometric sequence 14;y;4.

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two successive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2y(14)=4y

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for y.

    y(14)=4yy2=(4)(14)y2=1y=±1Don't forget the plus-minuswhich comes with the square root!y=±1

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 14;1;4 while if the constant ratio is negative the sequence will be 14;1;4.

    NOTE:

    While it does not say so, this question is about the geometric mean. If we have three terms in a geometric sequence, the term in the middle is the geometric mean of the other two. In this question, y is the geometric mean of 14 and 4, and it has two possible values: ±1.

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is positive. So we can ignore the 1 answer.

    The correct answer is 1.


    Submit your answer as:
  2. Find the general formula for the terms in the sequence in Question 1 if it is now given that the constant ratio of the sequence is 4.

    Answer: Tn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The answer will be an expression which includes the variable n.
    STEP: Use the formula Tn=arn1 to find the answer
    [−3 points ⇒ 0 / 3 points left]

    The question asks us to find the general formula for this sequence. The general formula is an expression which tells us how to calculate any term in the sequence.

    Start with the equation:

    Tn=arn1

    a represents the first term in the sequence, r represents the constant ratio, and n represents the rank (position) of the term in the sequence. For the sequence 14;1;4, the first term is a=14. The question tells us that the constant ratio is r=4. Substitute these numbers into the equation and simplify.

    Tn=(14)(4)n1=(14)(4)n(4)1=(14)(14)(4)n=(116)(4)n

    The general formula for the sequence is Tn=116(4)n.


    Submit your answer as:

The geometric mean

  1. Determine the value of p if the following three terms are the beginning of a geometric sequence:

    13;p;3,if p>0

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: p=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The number p is the geometric mean of 13 and 3. Use the constant ratio to write an equation which includes p. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the unknown term in the geometric sequence 13;p;3.

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two consecutive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2p(13)=3p

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for p.

    p(13)=3pp2=(3)(13)p2=1p=±1Don't forget the plus-minuswhich comes with the square root!p=±1

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 13;1;3 while if the constant ratio is negative the sequence will be 13;1;3.

    NOTE:

    While it does not say so, this question is about the geometric mean. If we have three terms in a geometric sequence, the term in the middle is the geometric mean of the other two. In this question, p is the geometric mean of 13 and 3, and it has two possible values: ±1.

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is positive. So we can ignore the 1 answer.

    The correct answer is 1.


    Submit your answer as:
  2. Find the common ratio for the three terms in Question 1.

    INSTRUCTION:If there is more than one answer, separate the answers by the ';' symbol.
    Answer: Common ratio =
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The common ratio is the quotient of any two successive terms in the sequence.
    STEP: Use two consecutive terms to calculate r
    [−3 points ⇒ 0 / 3 points left]

    We need to find the common ratio for the geometric sequence. We know from Question 1 that the sequence is 13;1;3.

    The constant ratio is the number (or expression) we need to multiply each term by to get the next term. We can use any two consecutive terms in the sequence to find the answer:

    r=T3T2orr=T2T1

    We can summarise this relationship as r=TnTn1. In this solution we will use the first and second terms.

    r=T2T1r=1(13)r=3

    The constant ratio is r=3.


    Submit your answer as:

The geometric mean

  1. What is the geometric mean of the following numbers if the mean is positive?

    13 and 127

    Give all possible answers.

    INSTRUCTION: If there is more than one answer, separate the answers by the ';' symbol.
    Answer: The geometric mean is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]
    The geometric mean of 13 and 127 is a number p which makes the numbers 13;p;127 follow a geometric pattern. Use the constant ratio to write an equation which includes p. Then solve the equation.
    STEP: Write an equation based on the constant ratio of the terms
    [−1 point ⇒ 1 / 2 points left]

    The question asks us to find the geometric mean for the numbers 13 and 127.

    This question is the same as: 'Find the value of p so that the terms 13;p;127 follow a geometric sequence.' This is because if 13;p;127 form a geometric sequence, then p is the geometric mean of 13 and 127. (If you have three consecutive terms of a geometric sequence, the middle term is always the geometric mean of the first and third terms.)

    Since we are dealing with a geometric sequence, we can find the answer using the constant ratio. Remember that the constant ratio is the quotient of any two consecutive terms in the sequence. We do not know the value of the constant ratio, but we do know that r is equal to both T2T1 and T3T2. So we can write the equation below with the first three terms and then substitute in the terms in this sequence:

    T2T1=T3T2p(13)=(127)p

    The equation above summarizes the relationship between the terms based on the constant ratio.


    STEP: Find the final answer or answers
    [−1 point ⇒ 0 / 2 points left]

    Now we can solve the equation for p.

    p(13)=(127)pp2=(127)(13)p2=181p=±181Don't forget the plus-minuswhich comes with the square root!p=±19

    We get two answers because of the ± which comes into the solution with the square-root step. This makes sense, because it is possible for the terms of a geometric sequence to change signs: if the constant ratio is positive, the sequence will be 13;19;127 while if the constant ratio is negative the sequence will be 13;19;127.

    NOTE:

    There is a formula for the geometric mean of two numbers: the geometric means of a and b are x=±ab. For more about the connection between this formula and the working shown above, see the explanation in the textbook.

    There are two answers at this point, but the question stated that the mean is positive. So we can ignore the 19 answer.

    The correct answer is 19.


    Submit your answer as:
  2. Find the general formula for the terms in the sequence in Question 1 if it is now given that the constant ratio of the sequence is 13.

    Answer: Tn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]
    The answer will be an expression which includes the variable n.
    STEP: Use the formula Tn=arn1 to find the answer
    [−3 points ⇒ 0 / 3 points left]

    The question asks us to find the general formula for this sequence. The general formula is an expression which tells us how to calculate any term in the sequence.

    Start with the equation:

    Tn=arn1

    a represents the first term in the sequence, r represents the constant ratio, and n represents the rank (position) of the term in the sequence. For the sequence 13;19;127, the first term is a=13. The question tells us that the constant ratio is r=13. Substitute these numbers into the equation and simplify.

    Tn=(13)(13)n1=(13)(13)n(13)1=(13)(31)(13)n=(13)n

    The general formula for the sequence is Tn=13n.


    Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

20;10;5;2,5;

Calculate the value of the fifth term, T5, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T5=
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the fifth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=20.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=1020=12

STEP: Calculate the value of the fifth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the fifth term.

T5=(20)(12)(51)=(20)(116)=1,25

Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

16;8;4;2;

Calculate the value of the sixth term, T6, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T6=
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the sixth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=16.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=816=12

STEP: Calculate the value of the sixth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the sixth term.

T6=(16)(12)(61)=(16)(132)=0,5

Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2.1
Maths formulas

The following geometric sequence is given:

16;8;4;2;

Calculate the value of the sixth term, T6, of this sequence.

INSTRUCTION: Your answer should be exact - do not round off.
Answer: T6=
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the formula for the general term of a geometric sequence.


STEP: Determine the values of a and r
[−1 point ⇒ 1 / 2 points left]

The formula for the general term of a geometric sequence is

Tn=arn1

Before we can use this formula to find the sixth term of the sequence, we need to know the values of a and r.

a is the first term of the sequence, so here a=16.

r is the constant ratio of the geometric sequence. We can find this by taking the ratio of any two consecutive terms in the sequence. Let's use the first and second terms:

r=T2T1=816=12

STEP: Calculate the value of the sixth term
[−1 point ⇒ 0 / 2 points left]

We can now use these values in the formula for the general term, to find the value of the sixth term.

T6=(16)(12)(61)=(16)(132)=0,5

Submit your answer as:

5. Geometric series

Exercises

6. Practical applications

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 20 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 24 cm, and the length of each successive rectangle is 90% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the seventh rectangle.

    Answer: The length is cm.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 90% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the seventh rectangle. The length of the first rectangle is a=24 cm and the constant ratio is r=0,9, so we can substitute and solve for Tn:

    Tn=arn1=(24)(0,9)6=12,75458...12,75 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 20 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=10 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(24)(1(0,9)10)(1(0,9))=(24)(0,65132...)(0,1)=156,31717...156,32 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=156,32 cm×1 cm=156,32 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=20×24=480 cm2

    So the percentage of the page that is coloured blue is:

    blue=156,32480×100%=32,56666...=32,57%

    Submit your answer as:

Testing for convergent and divergent series

  1. Determine the value of r for the following geometric series:

    2+(4)+8+(16)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=42=2Or r=T3T2=84=2

    Therefore, the constant ratio for this geometric series is r=2 .


    Submit your answer as:
  2. Does the given series diverge or converge?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=2, we can conclude that the series diverges.


    Submit your answer as:

Working with geometric series

A geometric series has the terms T4=27 and T5=81 and a sum of Sn=364.

  1. Determine the value of r for the series.

    Answer: The value of r is: .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−1 point ⇒ 0 / 1 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    Since T4 and T5 are consecutive terms, we can find r using the relationship r=TnTn1.

    r=TnTn1=T5T4=8127=3

    There we have it: the value of r is 3.


    Submit your answer as:
  2. How many terms are in the series? The first term of the series is T1=1.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r364=1(1(3)n)1(3)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    364=1(1(3)n)1(3)364=1(1(3)n)2(364)(2)=1(1(3)n)(728)(1)=1(3)n728=1(3)n7281=(3)n729=(3)n729=(3)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=5 to see if it agrees:

    (3)n(3)5=243

    That does not agree with the working above, because we need an answer of 729, not 243.

    If we continue to try different values for n, we will find that n=6 works:

    (3)n(3)6=729

    The correct answer for the number of terms in the series is n=6.


    Submit your answer as:

Using the general formula for the sum of a finite geometric series

The first term of a geometric series is 12 and the constant ratio is 2. If the series consists of 5 terms, determine the sum of the series.

Answer: The sum of the series =
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 5. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

12;1;2;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r>1, we will use the second form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(rn1)r1S5=12(251)21S5=312

Therefore, the sum of the geometric series is 312.


Submit your answer as:

Word problem: determine the maximum height

  1. A plant is 101 cm high and receives an average rainfall of 542 mm. After one year, the plant is 127 cm tall. For the years that follow, the growth of the plant is a third of the previous year's growth.

    Determine the maximum height to which the plant will grow. Give your answer rounded to 2 decimal places.

    Answer:

    Maximum height = cm

    numeric
    STEP: Write down a series for the annual growth of the plant
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the plant grows by 127 cm101 cm=26 cm in the first year. Therefore a=26. Be careful not to make the height (127 cm) the first term in the series. It is important to recognise that the series must describe the growth of the plant. We also need to calculate the constant ratio.

    26+263+269+2627+
    r=T2T1=26326=13Or r=T3T2=269263=13

    So we have that a=26 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the plant grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the plant has a limit or maximum.

    We determine the maximum growth of the plant by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=26113=2623=39

    Therefore, the growth of the plant is limited to 39 cm, and the maximum height of the plant is 101 cm+39 cm=140 cm.


    Submit your answer as:
  2. Write down the general term for the series.

    Answer:

    Tn=

    expression
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=26r=13
    Tn=arn1=26(13)n1

    Therefore, the general term for the series is given by Tn=26(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the plant is continuous.


    Submit your answer as:

Convergent series: determine the values of v

Calculate for which values of v the geometric series Tn=58n1(v+1)n1 will converge.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of v:

relational
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=5811(v+1)11=5For n=2:T2=5821(v+1)21=58(v+1)For n=3:T3=5831(v+1)31=582(v+1)2=564(v+1)2For n=4:T4=5841(v+1)41=583(v+1)3=5512(v+1)3

From the given series we can see that a=T1=5.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=58(v+1)5=18(v+1)Or r=T3T2=564(v+1)258(v+1)=18(v+1)

So we have that a=5 and r=18(v+1).


STEP: Apply the condition for convergence to determine the possible values of v
[−2 points ⇒ 0 / 4 points left]

For a geometric series to converge, the value of r must lie within the interval (1;1). We use this fact to determine the possible values of v:

1<r<11<18(v+1)<1Multiply through by 8:8<v+1<8Subtract 1:9<v<7

Therefore, for the series to converge, we have determined that 9<v<7.


Submit your answer as:

Finding the r and the Sn for a geometric series

The eighth term of a geometric series is 64. The tenth term is 256. Answer the questions below about the series.

  1. Determine the constant ratio for the terms in the series if the constant ratio is negative.

    Answer: The constant ratio is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−3 points ⇒ 0 / 3 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the eighth term is 64, which means T8=64. Similarly, the tenth term is 256, and we can write T10=256.

    T8 and T10 are not consecutive terms, so we cannot use the usual formula r=TnTn1 to find r. Instead, we will use the formula Tn=arn1. Based on this formula we can write T8=ar7 and T10=ar9. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T10T8 and then substitute in the expressions with a and r:

    T10T8=ar9ar725664=ar9ar7

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    4=r2±4=r2±2=r

    We get two answers from the calculation because of the "plus-minus" which comes in with the square root. However, the question told us that the constant ratio is negative. Therefore, we should throw out the positve answer. The correct answer for the constant ratio is r=2.


    Submit your answer as:
  2. Determine the sum of the first 9 terms. The first term of the series is T1=12.

    Answer:

    S9=

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 9 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS9=12(1(2)9)1(2)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S9=12(1(2)9)1(2)=12(1(512))3=12(513)3=(5132)(13)=1712

    The sum of the first 9 terms of the series is 1712.


    Submit your answer as:

Finding the sum to infinity of a geometric series

  1. Consider the following series and calculate r.

    8+(32)+128+(512)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=328=4Or r=T3T2=12832=4

    Therefore, we have that r=4.


    Submit your answer as:
  2. Find the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    one-of
    type(string.nocase)
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=4, we can conclude that the series diverges and sum to infinity does not exist.


    Submit your answer as:

Convergent series: determine the values of z

Determine for which values of z the geometric series Sn=2+26(z+1)+262(z+1)2 will be convergent.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of z: .
relational
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=26(z+1)2=16(z+1)Or r=T3T2=262(z+1)226(z+1)=16(z+1)

So we know that r=16(z+1).


STEP: Apply the condition for convergence to determine the possible values of z
[−2 points ⇒ 0 / 3 points left]

For a geometric series to converge, the value of r must lie within the interval (1;1). We use this fact to determine the possible values of z:

1<r<11<16(z+1)<1Multiply through by 6:6<z+1<6Subtract 1:7<z<5

We have the answer: for the series to converge, z must be in this range: 7<z<5.


Submit your answer as:

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

12;6;3;1,5;
  1. Complete the statement below about the following infinite series:

    12+6+3+1,5+
    Answer: The series converge because .
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=612=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=12 and r=12:

    S=a1r=12112=1212=24

    and for Sn:

    Sn=a(rn1)r1;r1=12((12)n1)121=12((12)n1)12=24((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=24(24((12)n1))=24+24(12)n24=24(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=24
    2. q=12

    Submit your answer as: and

Convergent and divergent series

  1. Given the general term Tn=(35)n, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by first writing down the first few terms in the series.


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=(35)n. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=(35)1=35For n=2:T2=(35)2=925For n=3:T3=(35)3=27125For n=4:T4=(35)4=81625

    Therefore, we have the following series:

    35+925+27125+81625+

    So, the first term for the series is a=35.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=92535=35Or r=T3T2=27125925=35

    Therefore, this is a geometric series with a=35 and r=35.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that 0<r<1, we use the following formula:

    Sn=a(1rn)1r

    We substitute in the known values and simplify:

    Sn=35(1(35)n)135=35(1(35)n)25=3525(1(35)n)=32(1(35)n)

    Therefore, the sum of the first n terms of the series is given by 32(1(35)n).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=32(1(35)n) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:(35)n01(35)n132(1(35)n)32Sn32

    Therefore, we know that the series is convergent because as n gets larger and larger, Sn tends to 32.


    Submit your answer as: and

Exercises

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 34 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 23 cm, and the length of each successive rectangle is 85% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the 12th rectangle.

    Answer: The length is cm.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 85% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the 12th rectangle. The length of the first rectangle is a=23 cm and the constant ratio is r=0,85, so we can substitute and solve for Tn:

    Tn=arn1=(23)(0,85)11=3,84889...3,85 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 34 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=17 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(23)(1(0,85)17)(1(0,85))=(23)(0,93688...)(0,15)=143,65594...143,66 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=143,66 cm×1 cm=143,66 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=34×23=782 cm2

    So the percentage of the page that is coloured blue is:

    blue=143,66782×100%=18,37084...=18,37%

    Submit your answer as:

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 26 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 20 cm, and the length of each successive rectangle is 90% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the eighth rectangle.

    Answer: The length is cm.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 90% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the eighth rectangle. The length of the first rectangle is a=20 cm and the constant ratio is r=0,9, so we can substitute and solve for Tn:

    Tn=arn1=(20)(0,9)7=9,56593...9,57 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 26 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=13 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(20)(1(0,9)13)(1(0,9))=(20)(0,74581...)(0,1)=149,16268...149,16 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=149,16 cm×1 cm=149,16 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=26×20=520 cm2

    So the percentage of the page that is coloured blue is:

    blue=149,16520×100%=28,68461...=28,68%

    Submit your answer as:

Stripes!

Adapted from DBE Nov 2016 Grade 12, P1, Q3.2
Maths formulas

Rectangles of width 1 cm are drawn from the edge of a sheet of paper that is 30 cm long, such that there is a 1 cm gap between one rectangle and the next. The length of the first rectangle is 24 cm, and the length of each successive rectangle is 85% of the length of the previous rectangle, until there are rectangles drawn along the entire length of AD. Each rectangle is coloured blue.

Answer the following questions about this scenario.

INSTRUCTION: Round all answers to two decimal places.
  1. Calculate the length of the seventh rectangle.

    Answer: The length is cm.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You may find it helpful to revise geometric sequences in the Everything Maths textbook.


    STEP: Use the general formula for a geometric sequence
    [−3 points ⇒ 0 / 3 points left]

    Each rectangle is 85% smaller than the one before it. Since there is a constant ratio between the lengths of the rectangles, we can use the general formula for a geometric sequence

    Tn=arn1

    to find the length of the seventh rectangle. The length of the first rectangle is a=24 cm and the constant ratio is r=0,85, so we can substitute and solve for Tn:

    Tn=arn1=(24)(0,85)6=9,05158...9,05 cm

    Submit your answer as:
  2. Calculate the percentage of the paper that is coloured blue.

    Answer:

    The percentage of paper that is coloured blue is %.

    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    You may find it helpful to revise finite geometric series in the Everything Maths textbook.


    STEP: Calculate the total area of the blue rectangles
    [−2 points ⇒ 2 / 4 points left]

    The lengths of the blue rectangles form a geometric series. So we can use the formula for the sum of a geometric series to calculate the total length of the rectangles on the page.

    Sn=a(1rn)(1r)

    The only value that we need to determine before we can use this calculation is the total number of rectangles on the page, n. Since the page is 30 cm high, and each rectangle has a height of 1 cm, with a space of 1 cm between each rectangle, there must be a total of n=15 rectangles going up the page.

    Substituting into the formula, we can calculate the sum of the lengths of the rectangles:

    Sn=a(1rn)(1r)=(24)(1(0,85)15)(1(0,85))=(24)(0,91264...)(0,15)=146,02332...146,02 cm

    Since the height of each of the rectangles is 1 cm, the total area of the rectangles is easy to calculate.

    A=146,02 cm×1 cm=146,02 cm2

    STEP: Calculate the percentage of paper that is covered
    [−2 points ⇒ 0 / 4 points left]

    The total area of the page is

    Apage=30×24=720 cm2

    So the percentage of the page that is coloured blue is:

    blue=146,02720×100%=20,28055...=20,28%

    Submit your answer as:

Testing for convergent and divergent series

  1. Consider the series shown below and calculate r.

    1+(1)+1+(1)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=11=1Or r=T3T2=11=1

    Therefore, the constant ratio for this geometric series is r=1 .


    Submit your answer as:
  2. Is the series convergent or divergent?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=1, we can conclude that the series diverges.


    Submit your answer as:

Testing for convergent and divergent series

  1. Consider the series shown below and calculate r.

    2+(2)+2+(2)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=22=1Or r=T3T2=22=1

    Therefore, the constant ratio for this geometric series is r=1 .


    Submit your answer as:
  2. Is this a divergent or convergent series?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=1, we can conclude that the series diverges.


    Submit your answer as:

Testing for convergent and divergent series

  1. Given the following geometric series, determine the value of the constant ratio r.

    (6)+(18)+(54)+(162)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−2 points ⇒ 0 / 2 points left]

    To calculate the constant ratio (r), we need to determine how consecutive terms in the series are related to each other:

    r=T2T1=186=3Or r=T3T2=5418=3

    Therefore, the constant ratio for this geometric series is r=3 .


    Submit your answer as:
  2. Is this a divergent or convergent series?

    Answer: The series .
    STEP: Use the test for convergence
    [−1 point ⇒ 0 / 1 points left]

    To test whether an infinte geometric series converges or diverges, we look at the value of the constant ratio:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=3, we can conclude that the series diverges.


    Submit your answer as:

Working with geometric series

A certain geometric series has a sum Sn=36481 and the following two terms: T2=1 and T5=127.

  1. Compute the value of r for the series.

    Answer: The value of r is: .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    T2 and T5 are not successive terms, so we need to put the formula Tn=arn1 to use. In particular, we know that T2=ar and T5=ar4. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T5T2 and then substitute in the expressions with a and r:

    T5T2=ar4ar(127)1=ar4ar

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    127=r33127=3r313=r

    The constant ratio is r=13.


    Submit your answer as:
  2. Compute the value of n. The first term of the series is T1=3.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r36481=3(1(13)n)1(13)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    36481=3(1(13)n)1(13)36481=3(1(13)n)23(36481)(23)=3(1(13)n)(728243)(13)=1(13)n728729=1(13)n7287291=(13)n1729=(13)n1729=(13)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=5 to see if it agrees:

    (13)n(13)5=1243

    That does not agree with the working above, because we need an answer of 1729, not 1243.

    If we continue to try different values for n, we will find that n=6 works:

    (13)n(13)6=1729

    The correct answer for the number of terms in the series is n=6.


    Submit your answer as:

Working with geometric series

Two terms in a geometric series, T4 and T5, have values of 9 and 27 respectively. The sum of the first n terms is Sn=613.

  1. Determine the value of r for the series.

    Answer: The value of r is: .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−1 point ⇒ 0 / 1 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    Since T4 and T5 are successive terms, we can find r using the relationship r=TnTn1.

    r=TnTn1=T5T4=279=3

    There we have it: the value of r is 3.


    Submit your answer as:
  2. How many terms are in the series? The first term of the series is T1=13.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r613=13(1(3)n)1(3)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    613=13(1(3)n)1(3)613=13(1(3)n)4(613)(4)=13(1(3)n)(2443)(3)=1(3)n244=1(3)n2441=(3)n243=(3)n243=(3)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=4 to see if it agrees:

    (3)n(3)4=81

    That does not agree with the working above, because we need an answer of 243, not 81.

    If we continue to try different values for n, we will find that n=5 works:

    (3)n(3)5=243

    The correct answer for the number of terms in the series is n=5.


    Submit your answer as:

Working with geometric series

Two terms in a geometric series, T1 and T4, have values of 1 and 8 respectively. The sum of the first n terms is Sn=85.

  1. Find the value of r for the terms in the series.

    Answer: The value of r is: .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We need to find the constant ratio, r, of the series. We have two terms in the series, and we can find r using them.

    T1 and T4 are not successive terms, so we need to put the formula Tn=arn1 to use. In particular, we know that T1=a and T4=ar3. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T4T1 and then substitute in the expressions with a and r:

    T4T1=ar3a81=ar3a

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    8=r338=3r32=r

    The constant ratio is r=2.


    Submit your answer as:
  2. Find the value of n.

    Answer: n=
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the sum given together with the formula Sn=a(1rn)1r


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 3 / 4 points left]

    We have the sum of the series, so it makes sense to use the formula Sn=a(1rn)1r in order to find the value of n. To get started, substitute all of the values we know into the formula.

    Sn=a(1rn)1r85=1(1(2)n)1(2)

    STEP: Solve the equation for n
    [−2 points ⇒ 1 / 4 points left]

    Now we need to rearrange the equation in order to find the value of n.

    85=1(1(2)n)1(2)85=1(1(2)n)3(85)(3)=1(1(2)n)(255)(1)=1(2)n255=1(2)n2551=(2)n256=(2)n256=(2)n

    STEP: Determine n by inspection
    [−1 point ⇒ 0 / 4 points left]

    At this point we must look for a value of n which agrees with the final line. For example, we can try the value n=7 to see if it agrees:

    (2)n(2)7=128

    That does not agree with the working above, because we need an answer of 256, not 128.

    If we continue to try different values for n, we will find that n=8 works:

    (2)n(2)8=256

    The correct answer for the number of terms in the series is n=8.


    Submit your answer as:

Using the general formula for the sum of a finite geometric series

You are provided with the following information regarding a geometric series:

a=14r=4n=6

Find its sum.

Answer: The sum of the series =
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 6. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

14;1;4;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r>1, we will use the second form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(rn1)r1S6=14(461)41S6=13654

Therefore, the sum of the geometric series is 13654.


Submit your answer as:

Using the general formula for the sum of a finite geometric series

The constant ratio of a geometric series consisting of 5 terms is 4. Given that the first term is 14, calculate the sum of the series.

Answer: The sum of the series =
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 5. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

14;1;4;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r<1, we will use the first form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(1rn)1rS5=14(1(4)5)1(4)S5=2054

Therefore, the sum of the geometric series is 2054.


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Using the general formula for the sum of a finite geometric series

You are provided with the following information regarding a geometric series:

a=13r=13n=8

Evaluate its sum.

Answer: The sum of the series =
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Notice that the geometric series consists of a specific number of terms. This means that you have a finite geometric series. Is there a general formula you can use for the sum of a finite geometric series? Do you have the needed values to use this formula?
STEP: Select the version of the general formula to use
[−1 point ⇒ 2 / 3 points left]

We are told in the question that there are a specific number of terms, 8. This means that we are dealing with a finite geometric series. Furthermore, we are given the first term and the constant ratio. We now have enough information to expand the series fully. The first three terms of our series are:

13;19;127;

However, writing out all the terms for a series can be tedious. Luckily, mathematicians have derived a general formula for calculating the sum of a finite geometric series. There are two forms of the general formula:

Sn=a(1rn)1r OR Sn=a(rn1)r1

The first version of the formula is easier to use when r<1 and the second when r>1.

As r<1, we will use the first form of the formula.


STEP: Substitute in the given values
[−2 points ⇒ 0 / 3 points left]

Our final step is to substitute the values given in the question into the appropriate general formula:

Sn=a(1rn)1rS8=13(1(13)8)1(13)S8=16406561

Therefore, the sum of the geometric series is 16406561.


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Word problem: determine the maximum height

  1. A bramble is 102 mm high and receives an average rainfall of 658 mm. After one year, the bramble is 137 mm tall. For the years that follow, the growth of the bramble is a third of the previous year's growth.

    Determine the maximum height to which the bramble will grow. Give your answer rounded to 2 decimal places.

    Answer:

    Maximum height = mm

    numeric
    STEP: Write down a series for the annual growth of the bramble
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the bramble grows by 137 mm102 mm=35 mm in the first year. Therefore a=35. Be careful not to make the height (137 mm) the first term in the series. It is important to recognise that the series must describe the growth of the bramble. We also need to calculate the constant ratio.

    35+353+359+3527+
    r=T2T1=35335=13Or r=T3T2=359353=13

    So we have that a=35 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the bramble grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the bramble has a limit or maximum.

    We determine the maximum growth of the bramble by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=35113=3523=52,5

    Therefore, the growth of the bramble is limited to 52,5 mm, and the maximum height of the bramble is 102 mm+52,5 mm=154,5 mm.


    Submit your answer as:
  2. Give the general term for the series.

    Answer:

    Tn=

    expression
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=35r=13
    Tn=arn1=35(13)n1

    Therefore, the general term for the series is given by Tn=35(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the bramble is continuous.


    Submit your answer as:

Word problem: determine the maximum height

  1. A weed is 76 mm high and receives an average rainfall of 681 mm. After one year, the weed is 110 mm tall. For the years that follow, the growth of the weed is a third of the previous year's growth.

    Determine the maximum height to which the weed will grow. Give your answer rounded to 1 decimal place.

    Answer:

    Maximum height = mm

    numeric
    STEP: Write down a series for the annual growth of the weed
    [−2 points ⇒ 2 / 4 points left]

    Read the question statement carefully. We notice that the initial height of the weed is 76 mm and it grows by 110 mm76 mm=34 mm in the first year. It is important to recognise that we need to write down a series that describes the growth of the weed. Therefore, a=34 and we must determine r.

    34+343+349+3427+
    r=T2T1=34334=13Or r=T3T2=349343=13

    So we have that a=34 and r=13. This was actually given to us in the question statement:

    third =13=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the weed grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the weed has a limit or maximum.

    We determine the maximum growth of the weed by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=34113=3423=51

    Therefore, the growth of the weed is limited to 51 mm, and the maximum height of the weed is 76 mm+51 mm=127 mm.


    Submit your answer as:
  2. Determine the general term for the series.

    Answer:

    Tn=

    expression
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=34r=13
    Tn=arn1=34(13)n1

    Therefore, the general term for the series is given by Tn=34(13)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the weed is continuous.


    Submit your answer as:

Word problem: determine the maximum height

  1. A hedge is 108 cm high and receives an average rainfall of 664 mm. After one year, the hedge is 123 cm tall. For the years that follow, the growth of the hedge is a quarter of the previous year's growth.

    Determine the maximum height to which the hedge will grow. Give your answer rounded to 1 decimal place.

    Answer:

    Maximum height = cm

    numeric
    STEP: Write down a series for the annual growth of the hedge
    [−2 points ⇒ 2 / 4 points left]

    From the question statement, we know that the hedge grows by 123 cm108 cm=15 cm in the first year. Therefore a=15. Be careful not to make the height (123 cm) the first term in the series. It is important to recognise that the series must describe the growth of the hedge. We also need to calculate the constant ratio.

    15+154+1516+1564+
    r=T2T1=15415=14Or r=T3T2=1516154=14

    So we have that a=15 and r=14. This was actually given to us in the question statement:

    quarter =14=r


    STEP: Apply the condition for convergence to determine the maximum height
    [−2 points ⇒ 0 / 4 points left]

    We notice that with each passing year, the hedge grows less and less. We also see that the constant ratio for this series lies within the interval (1;1), which means that the series will converge. In other words, the growth of the hedge has a limit or maximum.

    We determine the maximum growth of the hedge by calculating the value to which the series converges. To do that, we use the formula for the sum to infinity:

    S=a1r(r1)=15114=1534=20

    Therefore, the growth of the hedge is limited to 20 cm, and the maximum height of the hedge is 108 cm+20 cm=128 cm.


    Submit your answer as:
  2. Give the general term for the series.

    Answer:

    Tn=

    expression
    STEP: Write the general formula for the geometric series
    [−1 point ⇒ 0 / 1 points left]
    a=15r=14
    Tn=arn1=15(14)n1

    Therefore, the general term for the series is given by Tn=15(14)n1.

    Although it was not asked for in the question statement, a diagram is shown below to provide a visual representation of this series. It is important to note that we may join the points on the graph because the growth of the hedge is continuous.


    Submit your answer as:

Convergent series: determine the values of z

Calculate for which values of z the geometric series Tn=49n1(z1)n1 will converge.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of z:

relational
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=4911(z1)11=4For n=2:T2=4921(z1)21=49(z1)For n=3:T3=4931(z1)31=492(z1)2=481(z1)2For n=4:T4=4941(z1)41=493(z1)3=4729(z1)3

From the given series we can see that a=T1=4.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=49(z1)4=19(z1)Or r=T3T2=481(z1)249(z1)=19(z1)

So we have that a=4 and r=19(z1).


STEP: Apply the condition for convergence to determine the possible values of z
[−2 points ⇒ 0 / 4 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of z using:

1<r<11<19(z1)<1Multiply through by 9:9<z1<9Add 1:8<z<10

Therefore, for the series to converge, we have determined that 8<z<10.


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Convergent series: determine the values of w

Determine for which values of w the geometric series Tn=53n1(w+1)n1 will converge.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of w:

relational
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=5311(w+1)11=5For n=2:T2=5321(w+1)21=53(w+1)For n=3:T3=5331(w+1)31=532(w+1)2=59(w+1)2For n=4:T4=5341(w+1)41=533(w+1)3=527(w+1)3

From the given series we can see that a=T1=5.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=53(w+1)5=13(w+1)Or r=T3T2=59(w+1)253(w+1)=13(w+1)

So we have that a=5 and r=13(w+1).


STEP: Apply the condition for convergence to determine the possible values of w
[−2 points ⇒ 0 / 4 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of w using:

1<r<11<13(w+1)<1Multiply through by 3:3<w+1<3Subtract 1:4<w<2

Therefore, for the series to converge, we have determined that 4<w<2.


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Convergent series: determine the values of x

Determine for which values of x the geometric series Tn=27n1(x1)n1 will converge.
Give your answer as an inequality, for example: -1 < x < 1.

Answer:

Range of x:

relational
STEP: Determine the values of a and r
[−2 points ⇒ 2 / 4 points left]

Let's look at the first few terms in the series:

For n=1:T1=2711(x1)11=2For n=2:T2=2721(x1)21=27(x1)For n=3:T3=2731(x1)31=272(x1)2=249(x1)2For n=4:T4=2741(x1)41=273(x1)3=2343(x1)3

From the given series we can see that a=T1=2.

To determine the value of r, we need to determine the ratio between any two consecutive terms in the series:

r=T2T1=T3T2=TnTn1

r=T2T1=27(x1)2=17(x1)Or r=T3T2=249(x1)227(x1)=17(x1)

So we have that a=2 and r=17(x1).


STEP: Apply the condition for convergence to determine the possible values of x
[−2 points ⇒ 0 / 4 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of x using:

1<r<11<17(x1)<1Multiply through by 7:7<x1<7Add 1:6<x<8

Therefore, for the series to converge, we have determined that 6<x<8.


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Finding the r and the Sn for a geometric series

The fourth and fifth terms of a geometric series are 4 and 8 respectively. Answer the questions below about the series.

  1. Determine the constant ratio for the terms in the series.

    Answer: The constant ratio is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the fourth term is 4, which means T4=4. Similarly, the fifth term is 8, and we can write T5=8.

    T4 and T5 are successive terms so we can find r using the relationship r=TnTn1.

    r=TnTn1=T5T4=84=2

    There we have the answer: the constant ratio is 2.


    Submit your answer as:
  2. Determine the sum of the first 9 terms. The first term of the series is T1=12.

    Answer:

    S9=

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 9 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS9=12(1(2)9)1(2)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S9=12(1(2)9)1(2)=12(1(512))1=12(511)1=(5112)(1)=5112

    The sum of the first 9 terms of the series is 5112.


    Submit your answer as:

Finding the r and the Sn for a geometric series

The seventh and eighth terms of a geometric series are 5103 and 15309 respectively. Answer the questions below about the series.

  1. Find the value of r for the series.

    Answer: The value of r is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−2 points ⇒ 0 / 2 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the seventh term is 5103, which means T7=5103. Similarly, the eighth term is 15309, and we can write T8=15309.

    T7 and T8 are consecutive terms so we can find r using the relationship r=TnTn1.

    r=TnTn1=T8T7=153095103=3

    There we have the answer: the value of r is 3.


    Submit your answer as:
  2. Find the sum of the first 5 terms. The first term of the series is T1=7.

    Answer:

    S5=

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 5 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS5=7(1(3)5)1(3)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S5=7(1(3)5)1(3)=7(1(243))2=7(242)2=16942=847

    The sum of the first 5 terms of the series is 847.


    Submit your answer as:

Finding the r and the Sn for a geometric series

The sixth and ninth terms of a geometric series are 116 and 1128 respectively. Answer the questions below about the series.

  1. Compute the constant ratio for the terms in the series.

    Answer: The constant ratio is:
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the two terms given to figure out what the constant ratio, r, is for the series.


    STEP: Use the terms given to calculate the constant ratio
    [−3 points ⇒ 0 / 3 points left]

    We have the values for two terms in a geometric series, and we need to find the constant ratio, r, of the series. To begin, it will be helpful to write down what we know: the sixth term is 116, which means T6=116. Similarly, the ninth term is 1128, and we can write T9=1128.

    T6 and T9 are not successive terms, so we cannot use the usual formula r=TnTn1 to find r. Instead, we will use the formula Tn=arn1. Based on this formula we can write T6=ar5 and T9=ar8. Now we can create a ratio of these two expressions in order to find r. Start with the ratio T9T6 and then substitute in the expressions with a and r:

    T9T6=ar8ar5(1128)(116)=ar8ar5

    Now we can simplify the fractions on both sides of the equation. Then solve for r.

    18=r3318=3r312=r

    The constant ratio is r=12.


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  2. Compute the sum of the first 7 terms. The first term of the series is T1=2.

    Answer:

    S7=

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Substitute a, r and n into the sum formula Sn=a(1rn)1r.


    STEP: Substitute all known values into the formula Sn=a(1rn)1r
    [−1 point ⇒ 2 / 3 points left]

    We can use the formula Sn=a(1rn)1r to find the sum of 7 terms. Substitute all the values we know into the formula.

    Sn=a(1rn)1rS7=2(1(12)7)1(12)

    STEP: Evaluate the formula to get the sum
    [−2 points ⇒ 0 / 3 points left]

    Evaluate the right hand side of the formula. Start inside of the brackets in the numerator and also by evaluating the denominator.

    S7=2(1(12)7)1(12)=2(1(1128))32=2(129128)32=(12964)(23)=4332

    The sum of the first 7 terms of the series is 4332.


    Submit your answer as:

Finding the sum to infinity of a geometric series

  1. Determine the value of r for the given series:

    (32)+98+(2732)+81128+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=9832=34Or r=T3T2=273298=34

    Therefore, we have that r=34.


    Submit your answer as:
  2. Find the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    numeric
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=34, we can conclude that the series converges and sum to infinity does exist.

    We write down the formula for the sum to infinity S and substitute the values for a and r:

    a=32 andr=34S=a1r=321(34)=67
    Therefore, as the number of terms in the series tends towards infinity, the sum of the series tends to 67. This means that no matter how many terms are added to the series, the value of the sum will never be greater than 67.


    Submit your answer as:

Finding the sum to infinity of a geometric series

  1. Consider the following series and calculate r.

    1+(12)+14+(18)+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=121=12Or r=T3T2=1412=12

    Therefore, we have that r=12.


    Submit your answer as:
  2. Calculate the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    numeric
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=12, we can conclude that the series converges and sum to infinity does exist.

    We write down the formula for the sum to infinity S and substitute the values for a and r:

    a=1 andr=12S=a1r=11(12)=23
    Therefore, as the number of terms in the series tends towards infinity, the sum of the series tends to 23. This means that no matter how many terms are added to the series, the value of the sum will never be greater than 23.


    Submit your answer as:

Finding the sum to infinity of a geometric series

  1. Consider the following series and calculate r.

    (23)+49+(827)+1681+
    Answer: r=
    numeric
    STEP: Determine the value of r
    [−1 point ⇒ 0 / 1 points left]

    To calculate r, we need to determine the ratio between any two consecutive terms in the series:

    r=T2T1=4923=23Or r=T3T2=82749=23

    Therefore, we have that r=23.


    Submit your answer as:
  2. Determine the sum to infinity.

    INSTRUCTION: If it does not exist, write "Not exist" in the input box.
    Answer: S=
    numeric
    STEP: Use the test for convergence to determine if the sum to infinity exists
    [−2 points ⇒ 0 / 2 points left]

    We need to look at the value of the constant ratio to determine whether an infinte geometric series converges or diverges:

    • If 1<r<1, then the infinite geometric series converges.
    • If r<1 or r>1, then the infinite geometric series diverges.

    Therefore, since r=23, we can conclude that the series converges and sum to infinity does exist.

    We write down the formula for the sum to infinity S and substitute the values for a and r:

    a=23 andr=23S=a1r=231(23)=25
    Therefore, as the number of terms in the series tends towards infinity, the sum of the series tends to 25. This means that no matter how many terms are added to the series, the value of the sum will never be greater than 25.


    Submit your answer as:

Convergent series: determine the values of x

Calculate for which values of x the geometric series Sn=5+59(x1)+592(x1)2 will converge.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of x: .
relational
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=59(x1)5=19(x1)Or r=T3T2=592(x1)259(x1)=19(x1)

So we know that r=19(x1).


STEP: Apply the condition for convergence to determine the possible values of x
[−2 points ⇒ 0 / 3 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of x using:

1<r<11<19(x1)<1Multiply through by 9:9<x1<9Add 1:8<x<10

We have the answer: for the series to converge, x must be in this range: 8<x<10.


Submit your answer as:

Convergent series: determine the values of x

Calculate for which values of x the geometric series Sn=4+49(x+1)+492(x+1)2 will converge.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of x: .
relational
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=49(x+1)4=19(x+1)Or r=T3T2=492(x+1)249(x+1)=19(x+1)

So we know that r=19(x+1).


STEP: Apply the condition for convergence to determine the possible values of x
[−2 points ⇒ 0 / 3 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of x using:

1<r<11<19(x+1)<1Multiply through by 9:9<x+1<9Subtract 1:10<x<8

We have the answer: for the series to converge, x must be in this range: 10<x<8.


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Convergent series: determine the values of z

Calculate for which values of z the geometric series Sn=3+36(z1)+362(z1)2 will converge.

INSTRUCTION: Give your answer as an inequality.
Answer: Range of z: .
relational
STEP: Determine the value of r
[−1 point ⇒ 2 / 3 points left]

A geometric series converges (has a finite sum) if the constant ratio is between 1 and 1. So we need to determine the value of r for this series. As always, the ratio comes from any two consecutive terms:

r=T2T1=T3T2=TnTn1

r=T2T1=36(z1)3=16(z1)Or r=T3T2=362(z1)236(z1)=16(z1)

So we know that r=16(z1).


STEP: Apply the condition for convergence to determine the possible values of z
[−2 points ⇒ 0 / 3 points left]

The sum to infinity only exists for a converging geometric series. We know that a geometric series converges if 1<r<1, so we can determine the possible values of z using:

1<r<11<16(z1)<1Multiply through by 6:6<z1<6Add 1:5<z<7

We have the answer: for the series to converge, z must be in this range: 5<z<7.


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Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

20;10;5;2,5;
  1. Complete the statement below about the following infinite series:

    20+10+5+2,5+
    Answer: The series converge because .
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=1020=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=20 and r=12:

    S=a1r=20112=2012=40

    and for Sn:

    Sn=a(rn1)r1;r1=20((12)n1)121=20((12)n1)12=40((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=40(40((12)n1))=40+40(12)n40=40(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=40
    2. q=12

    Submit your answer as: and

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

8;4;2;1;
  1. Complete the statement below about the following infinite series:

    8+4+2+1+
    Answer: The series converge because .
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=48=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=8 and r=12:

    S=a1r=8112=812=16

    and for Sn:

    Sn=a(rn1)r1;r1=8((12)n1)121=8((12)n1)12=16((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=16(16((12)n1))=16+16(12)n16=16(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=16
    2. q=12

    Submit your answer as: and

Geometric sequences

Adapted from DBE Nov 2015 Grade 12, P1, Q2
Maths formulas

Given the geometric sequence:

14;7;3,5;1,75;
  1. Complete the statement below about the following infinite series:

    14+7+3,5+1,75+
    Answer: The series converge because .
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You may find it helpful to revise infinite series in the Everything Maths textbook.


    STEP: Test the series against conditions for convergence
    [−2 points ⇒ 0 / 2 points left]

    Whether or not a geometric series will converge depends on the value of the constant ratio r. If 1<r<1, the series will converge and the sum of the terms will tend towards some finite value. Otherwise, the sum of the terms will be infinite.

    We work out the value of r by taking the ratio of any two consecutive terms in the sequence.

    r=T2T1=714=12

    This value of r is 1<12<1.

    So the series will converge because 1<r<1.


    Submit your answer as: and
  2. We can find an expression for SSn in the form pqn, where Sn is the sum of the first n terms of the given geometric sequence. What are the values of p and q?

    INSTRUCTION: Type the values of p and q into the boxes below. Do not include any brackets.
    Answer:
    1. p=
    2. q=
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Determine the values of S and Sn, and then simplify to find an expression of the form abn.


    STEP: Determine the values of S and Sn
    [−2 points ⇒ 2 / 4 points left]
    S=a1r;1<r<1

    Substitute in the values a=14 and r=12:

    S=a1r=14112=1412=28

    and for Sn:

    Sn=a(rn1)r1;r1=14((12)n1)121=14((12)n1)12=28((12)n1)

    STEP: Determine the values of p and q
    [−2 points ⇒ 0 / 4 points left]

    We can now subtract Sn from S and simplify:

    SSn=28(28((12)n1))=28+28(12)n28=28(12)n

    This is in the correct form pqn. So the correct answers are:

    1. p=28
    2. q=12

    Submit your answer as: and

Convergent and divergent series

  1. Consider the general term Tn=4×5n+1, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Start by first writing down the first few terms in the series.


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=4×5n+1. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=4×52=100For n=2:T2=4×53=500For n=3:T3=4×54=2500For n=4:T4=4×55=12500

    Therefore, we have the following series:

    100+500+2500+12500+

    So, the first term for the series is a=100.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=500100=5Or r=T3T2=2500500=5

    Therefore, this is a geometric series with a=100 and r=5.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that r>1, we use the following formula:

    Sn=a(rn1)r1

    We substitute in the known values and simplify:

    Sn=100(5n1)51=100(5n1)4=1004(5n1)=25(5n1)

    Therefore, the sum of the first n terms of the series is given by 25(5n1).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=25(5n1) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:5n5n125(5n1)Sn

    Therefore, we know that the series is divergent because as n gets larger and larger, Sn tends to infinity.


    Submit your answer as: and

Convergent and divergent series

  1. Consider the general term Tn=(14)n, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Is this an arithmetic or geometric series?


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Let's look at the general term Tn=(14)n. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=(14)1=14For n=2:T2=(14)2=116For n=3:T3=(14)3=164For n=4:T4=(14)4=1256

    Therefore, we have the following series:

    14+116+164+1256+

    So, the first term for the series is a=14.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=11614=14Or r=T3T2=164116=14

    Therefore, this is a geometric series with a=14 and r=14.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that 0<r<1, we use the following formula:

    Sn=a(1rn)1r

    We substitute in the known values and simplify:

    Sn=14(1(14)n)114=14(1(14)n)34=1434(1(14)n)=13(1(14)n)

    Therefore, the sum of the first n terms of the series is given by 13(1(14)n).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=13(1(14)n) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:(14)n01(14)n113(1(14)n)13Sn13

    Therefore, we know that the series is convergent because Sn13 as n.


    Submit your answer as: and

Convergent and divergent series

  1. Given the general term Tn=4n, determine an expression for the sum of the first n terms of the series.

    Answer: Sn=
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    What kind of series does this general term give us?


    STEP: Use the general term to write down the first few terms in the series
    [−2 points ⇒ 1 / 3 points left]

    Consider the general term Tn=4n. We need to determine if this will generate an arithmetic or geometric series, so we write down the first few terms in the series:

    For n=1:T1=41=4For n=2:T2=42=16For n=3:T3=43=64For n=4:T4=44=256

    Therefore, we have the following series:

    4+16+64+256+

    So, the first term for the series is a=4.

    Now we need to see if there is a constant ratio (r), which would make this a geometric series:

    r=T2T1=164=4Or r=T3T2=6416=4

    Therefore, this is a geometric series with a=4 and r=4.


    STEP: Determine an expression for the sum of the first n terms of the series
    [−1 point ⇒ 0 / 3 points left]

    Since we know that we are dealing with a geometric series and that r>1, we use the following formula:

    Sn=a(rn1)r1

    We substitute in the known values and simplify:

    Sn=4(4n1)41=4(4n1)3=43(4n1)

    Therefore, the sum of the first n terms of the series is given by 43(4n1).


    Submit your answer as:
  2. Is the series convergent or divergent? Use the expression for the sum of the first n terms of the series to complete the following sentence:

    Answer:

    We can conclude that the given series is because:

    STEP: Consider the sum to infinity
    [−2 points ⇒ 0 / 2 points left]

    To complete the sentence given in the question statement, we look at the expression Sn=43(4n1) more closely and consider what happens to the sum as n tends to infinity.

    Let's break the expression down a bit:

    As n:4n4n143(4n1)Sn

    Therefore, we know that the series is divergent because as n gets larger and larger, Sn tends to infinity.


    Submit your answer as: and